I posted this proof on AoPS but I'm having trouble understanding what's wrong with it. I was hoping this community could offer a different perspective?
Suppose the claim is false. That is to say, suppose there exists an integer pair $(a,b)$ such that $2^b-1$ evenly divides $2^a+1$. For a given $b_0>2$ there exists a non-empty set of positive integers $S=\left\{a_0,\ldots,a_n\right\}$ that are counterexamples to the claim (we let $b_0>2$ because $(a,1)$ and $(3,2)$ are counterexamples). Let $a_0$ be the smallest value of $S$. Then $$\frac{2^{a_0} +1}{2^{b_0}-1}=k$$for some $k\in \mathbb{Z}$. Therefore, $$\frac{2^{a_0}+1+2^{b_0}-1}{2^{b_0}-1}=k+1$$Hence, $$\frac{2^{a_0}+2^{b_0}}{2^{b_0}-1}=k+1$$So, $$\frac{2^{b_0}(2^{a_0-b_0}+1)}{2^{b_0}-1}=k+1$$Because $2^{b_0}-1\nmid 2^{b_0}$, we find that $2^{b_0}-1\mid 2^{a_0-b_0}+1$, so $a_0-b_0\in S$. However, $a_0-b_0<a_0,$ contradicting the minimality of $a_0$. Therefore, $S$ is empty. $\blacksquare$