Which natural numbers $m$ and $n$ satisfy the equation $2^{n}+1=m^{2}$?
As $2^{n}=m^{2}-1=(m+1)(m-1)$, we establish that $m-1$ and $m+1$ are powers of 2. Since the difference between $m+1$ and $m-1$ is 2, the only possible solution is $m-1=2$ and $m+1=2^{2}$, from which $m=3$. Thus, $2^{n}+1=3^{2}=9$ and we get $n=3$. The answer is $m=n=3$.