##JavaScript (ES6), 96 bytes
A full program that prompts for the number of matching integers and displays them one by one, using alert().
for(i=prompt(n=2);i;n+=2)(g=b=>b>n?alert(n,i--):(C=(n,x=n)=>n%--x?C(n,x):x>1)(n^b|1)&&g(b*2))(1)Unless your browser is set up to use Tail Call Optimization, this will eventually break because of a recursion overflow.
Below is a non-recursive version (102 bytes).
for(i=prompt(n=2);i;n+=2){for(c=b=1;b<n;b*=2,c&=C)for(C=k=2,x=n^b|1;k<x;k++)C|=!(x%k);c&&alert(n,i--)}
Assumption
This algorithm relies on the assumption that all bitflip-resistant composite numbers are even. This leads to a rather important simplification: instead of flipping every possible pair of bits, we only flip bit #0 and another one (or no other bit at all) and check that all resulting numbers are composite.
However, I can't figure out any obvious proof that an odd bitflip-resistant composite number doesn't actually exist. It just happens to never be the case for small numbers (I checked up to 1,000,000), and it seems like the probability of finding one is decreasing as the number of bits is increasing (but this is basically just my intuition about it).