Which natural numbers $m$ and $n$ satisfy $2^{n}+1=m^{2}$?
As

$$
2^{n}=m^{2}-1=(m+1)(m-1)
$$

we have that $m-1$ and $m+1$ are powers of 2 whose difference is 2. Therefore, the only possible solution is $m-1=2$ and $m+1=2^{2}$, from which $m=3$. It follows that $2^{n}+1=3^{2}$, and we obtain $n=3$.