We have $2^n \mid (2a+1)^n + (2b+1)^n$. Note that $n=1$ works clearly, suppose $n\ge 2$. We first show $n$ is odd. Assume $n$ is even. Then, $4\mid (2a+1)^n + (2b+1)^n$, whereas $(2a+1)^n\equiv (2b+1)^n\equiv 1\pmod{4}$, a contradiction. Having shown this, we arrive at \[ 2^n \mid (2a+2b+2)\bigl((2a+1)^{n-1} + (2a+1)^{n-2}(2b+1)+\cdots+(2b+1)^{n-1}\bigr). \]As $n$ is odd, the second term above is odd, yielding that $2^{n-1}\mid a+b+1$. Since $a,b$ are fixed, we must therefore have $n\le v_2(a+b+1)+1$, thus the set of all such $n$ is indeed finite.