Find all integers $b, n \in \mathbb{N}$ such that $b^{2}-8=3^{n}$.
For $n=0$, it involves solving $b^{2}-8=1$ thus $b^{2}=9$ and $b=3$.

If $(b, n)$ is another solution with $n \geq 1$, we have $b^{2}=3^{n}+8 \equiv 8 \equiv 2[3]$. However, a square is never congruent to 2 modulo 3 (the square of 0 is 0, that of 1 is 1, that of 2 is $2^{2}=4 \equiv 1$).