(French-Slovak Competition 1996) Find all strictly positive integers $x, y, p$ such that $p^{x}-y^{p}=1$ with $p$ prime.
Let's rewrite the equation in the form $y^{p}+1^{p}=p^{x}$. If $y=1$, we see that $p=2$ and $x=1$. If $p=2$, it readily follows that necessarily $x, y=1$. Therefore, we assume $p$ is odd so that $y+1$ divides $y^{p}+1$.

If the conditions of Zsigmondy's theorem are satisfied, there exists a prime factor of $y^{p}+1$ that does not divide $y+1$, so $y^{p}+1$ cannot be a power of a prime number. It remains to handle the case $(y, p)=(2,3)$, which gives the solution $x=2$.