$(x-y)(x+y)=2^m$ $GCD(x-y,x+y)<=2y$ $2y|2^m$ so $y=2^k$ Now if we decide with $2^m$we have : $2^a-1=b^2$ If $a>=2$ then mod4 gives a contradiction. So $a=1$ and $b=1$.

jasperE3 wrote: Let $S$ be the set of positive integers $x$ for which there exist positive integers $y$ and $m$ such that $y^2-2^m=x^2$. (a) Find all of the elements of $S$. (b) Find all $x$ such that both $x$ and $x+1$ are in $S$. very nice problem! For any $y$ where is solution ?