Find all natural numbers $m, n$ such that $m^{2}-8=3^{n}$.
First, suppose $n=0$. We then have $3^{n}=1$, which gives $m^{2}=9$ and thus $m=3$ since $m$ is positive. We find the solution $m=3, n=0$, and we verify that it works.

Now suppose $n>0$. We then have 3 dividing $3^{n}$. By looking modulo 3, we get $m^{2}-8 \equiv 0(\bmod 3)$. Since $8 \equiv 2(\bmod 3)$, we have $m^{2} \equiv 2(\bmod 3)$. A simple calculation of squares modulo 3 shows that this equation has no other solution.