Pyth, 9 bytes

• Borrowing the x & (x*2) != 0 algorithm from @alephalpha
• Borrowing Pyth boilerplate from @Jakube

My first attempt at Pyth:

f!.&TyThQ

Try it here

               implicit: Q = input()            
f      hQ      find the first integer T >= Q + 1, 
               that satisfies the condition:
 !.&TyT        T & (T * 2) is 0

CJam, 14 11 bytes

3 bytes saved thanks to DigitalTrauma.

l~{)___+&}g

Test it here.

Explanation

l~          "Read and eval input.";
  {      }g "Do while...";
   )_       "Increment and duplicate (call this x).";
     __+    "Get two more copies and add them to get x and 2x on the stack.";
        &   "Take their bitwise AND. This is non-zero is as long as x's base-2
             representation contains '11'.";

This leaves the last number on the stack which is printed automatically at the end of the program.

Python 2, 44 bytes

This is a complete python program that reads in n, and prints the answer. I think it does quite well in the readability sub-competition.

n=input()+1
while'11'in bin(n):n+=1
print n

The test results:

$ echo 12 | python soln.py 
16
$ echo 18 | python soln.py 
20

Pyth, 12 11 bytes

f!}`11.BThQ

Try it online: Pyth Compiler/Executor.

               implicit: Q = input()            
f        hQ    find the first integer T >= Q + 1, 
               that satisfies the condition:
 !}`11.BT         "11" is not in the binary representation of T

Mathematica, 41 30 bytes

Saved 11 bytes thanks to Martin Büttner.

#+1//.i_/;BitAnd[i,2i]>0:>i+1&

Perl, 31

#!perl -p
sprintf("%b",++$_)=~/11/&&redo

Or from the command line:

 perl -pe'sprintf("%b",++$_)=~/11/&&redo' <<<"18"

APL, 18 bytes

1∘+⍣{~∨/2∧/⍺⊤⍨⍺⍴2}

This evaluates to a monadic function. Try it here. Usage:

   1∘+⍣{~∨/2∧/⍺⊤⍨⍺⍴2} 12
16

Explanation

1∘+                    ⍝ Increment the input ⍺
   ⍣{            }     ⍝ until
     ~∨/               ⍝ none of
        2∧/            ⍝ the adjacent coordinates contain 1 1 in
           ⍺⊤⍨⍺⍴2      ⍝ the length-⍺ binary representation of ⍺.

J, 20 characters

A monadic verb. Fixed to obey the rules.

(+1 1+./@E.#:)^:_@>:

Explanation

First, this is the verb with spaces and then a little bit less golfed:

(+ 1 1 +./@E. #:)^:_@>:
[: (] + [: +./ 1 1 E. #:)^:_ >:

Read:

    ]                             The argument
      +                           plus
        [: +./                    the or-reduction of
               1 1 E.             the 1 1 interval membership in
                      #:          the base-2 representation of the argument,
[: (                    )^:_      that to the power limit of
                             >:   the incremented argument

The argument plus the or-reduction of the 1 1 interval membership in the base-2 representation of the argument, that to the power limit applied to the incremented argument.

I basically compute if 1 1 occurs in the base-2 representation of the input. If it does, I increment the input. This is put under a power-limit, which means that it is applied until the result doesn't change any more.

Javascript, 25 19

Using the fact that, for a 1-sparse binary number, x&2*x == 0:

f=x=>x++&2*x?f(x):x

JavaScript (ES6), 39 43

No regexp, no strings, recursive:

R=(n,x=3)=>x%4>2?R(++n,n):x?R(n,x>>1):n

Iterative version:

F=n=>{for(x=3;x%4>2?x=++n:x>>=1;);return n}

It's very simple, just using right shift to find a sequence of 11. When I find it, skip to next number. The recursive version is directly derived from the iterative one.

Ungolfed and more obvious. To golf, the trickiest part is merging the inner and outer loops (having to init x to 3 at start)

F = n=>{
  do {
    ++n; // next number
    for(x = n; x != 0; x >>= 1) {
      // loop to find 11 in any position
      if ((x & 3) == 3) { // least 2 bits == 11
        break;
      }
    }
  } while (x != 0) // if 11 was found,early exit from inner loop and x != 0
  return n
}

Python 2, 37 bytes

f=input()+1
while f&2*f:f+=1
print f

Used the logic x & 2*x == 0 for 1-sparse number.
Thanks to @Nick and @CarpetPython.

JavaScript, 75 66 62 bytes

Thanks to Martin Büttner for saving 9 bytes and Pietu1998 for 4 bytes!

function n(a){for(a++;/11/.test(a.toString(2));a++);return a;}

How it works: it runs a for loop starting from a + 1 as long as the current number is not 1-sparse, and if it is, the loop is interrupted and it returns the current number. To check whether a number is 1-sparse, it converts it to binary and checks whether it does not contain 11.

Un-golfed code:

function nextOneSparseNumber(num) {
    for (num++; /11/.test(num.toString(2)); num++);
    return num;
}

Julia, 40 bytes

n->(while contains(bin(n+=1),"11")end;n)

This creates an anonymous function that accepts a single integer as input and returns the next highest 1-sparse integer. To call it, give it a name, e.g. f=n->..., and do f(12).

Ungolfed + explanation:

function f(n)

    # While the string representation of n+1 in binary contains "11",
    # increment n. Once it doesn't, we've got the answer!

    while contains(bin(n += 1), "11")
    end

    return(n)
end

Examples:

julia> f(12)
16

julia> f(16)
20

Suggestions and/or questions are welcome as always!

Python, 39 33 bytes

Try it here: http://repl.it/gpu/2

In lambda form (thanks to xnor for golfing):

f=lambda x:1+x&x/2and f(x+1)or-~x

Standard function syntax turned out to be shorter than a lambda for once!

def f(x):x+=1;return x*(x&x*2<1)or f(x)

><> (Fish), 31 + 3 = 34 bytes

1+:>:  4%:3(?v~~
;n~^?-1:,2-%2<

Usage:

>python fish.py onesparse.fish -v 12
16

3 bytes added for the -v flag.

JavaScript (ECMAScript 6), 40

By recursion:

g=x=>/11/.test((++x).toString(2))?g(x):x

JavaScript, 56

Same without arrow functions.

function f(x){return/11/.test((++x).toString(2))?f(x):x}

Scala, 65 bytes

(n:Int)=>{var m=n+1;while(m.toBinaryString.contains("11"))m+=1;m}

(if a named function is required, solution will be 69 bytes)

Perl, 16 bytes

Combining the x&2*x from various answers (I think Nick's first) with nutki's redo yields:

perl -pe'++$_&2*$_&&redo'

Tested in Strawberry 5.26.

Japt, 8 bytes

_&ZÑ}fUÄ

Run it online.

Stax, 5 bytes

╦>ù╤x

Run and debug it

It works using this procedure. The input starts on top of the stack.

• Increment and copy twice.
• Halve the top of the stack.
• Bitwise-and top two elements on the stack.
• If the result is truthy (non-zero) repeat the entire program.

Java, 33 bytes.

Uses the method in this answer

n->{for(;(n++&2*n)>0;);return n;}

TIO

Vyxal, 6 bytes

λd⋏[›x

Try it Online!

How?

λd⋏[›x
λ      # Open a lambda for recursion
 d     #  Double
  ⋏    #  Bitwise AND the doubled input with the input
   [   #  If this is truthy (non-zero):
    ›x #   Increment and recurse

Jelly,  7  6 bytes

‘‘&Ḥ$¿

A monadic Link that accepts a non-negative integer and yields a (higher) positive integer, the next higher 1-sparse integer.

Try it online!

How?

Start with v=n+1. Then increment v if doubling v to shift every bit up one place and then bit-wise AND-ing that with v is non-zero (meaning v is not 1-sparse); repeat until we hit a 1-sparse number.

‘‘&Ḥ$¿ - Main Link: n   e.g. 12
‘      - increment -> v=n+1  13
     ¿ - while...
    $  - ...condition: last two links as a monad - f(v):
   Ḥ   -                 double v -> 2v
  &    -                 (v) bit-wise AND (2v) -> is v not 1-sparse?
 ‘     - ...do: increment

Vyxal 3, 7 bytes

λ:d∴[ꜝx

Vyxal It Online!

Uses the same algorithm as the Vyxal 2 answer.

Alternate algorithm for 14 bytes: (Vyxal It Online!)

ÞṆİ:dZ∴Ṡᶤ0=⁰+ꜝ­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏⁠‎⁡⁠⁣⁢‏⁠‎⁡⁠⁣⁣‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁣⁤‏⁠‎⁡⁠⁤⁡‏⁠‎⁡⁠⁤⁢‏‏​⁡⁠⁡‌­
ÞṆİ             # ‎⁡For every positive integer after the input:
   :dZ∴         # ‎⁢Bitwise AND it with itself doubled
       Ṡᶤ0=     # ‎⁣Find the index of the first result which is equal to zero
           ⁰+ꜝ  # ‎⁤and add the input to that index.
💎

Created with the help of Luminespire.

Ruby, 44

->(i){loop{i+=1;break if i.to_s(2)!~/11/};i}

Pretty basic. A lambda with an infinite loop and a regexp to test the binary representation. I wish that loop yielded and index number.

Matlab (77 74 bytes)

m=input('');for N=m+1:2*m
if ~any(regexp(dec2bin(N),'11'))
break
end
end
N

Notes:

• It suffices to test numbers m+1 to 2*m, where m is the input.
• ~any(x) is true if x contains all zeros or if x is empty

C (32 bytes)

f(int x){return 2*++x&x?f(x):x;}

Recursive implementation of the same algorithm as so many other answers.