Find all integers $x, y \geq 0$ such that $2^{x}=3^{y}+5$.
For $x \leq 5, y \leq 3$, we find the solutions $(x, y)=(3,1)$ and $(x, y)=$ $(5,3)$. We assume therefore $x \geq 6$ and $y \geq 4$.

It is natural to start by looking modulo $3^{4}, 3^{5}, \ldots, 2^{6}, 2^{7}, \ldots$. We look modulo $2^{6}$: the order of 3 modulo 64 is 16. We find that $y \equiv 11 \bmod 16$. To make $3^{y}$ constant, it is natural to look modulo $p=17$, because then $3^{y}+5 \equiv 3^{11}+5 \equiv 12 \bmod 17$. However, we verify that the powers of 2 are never 12 modulo 17. This shows that there are no other solutions.