(USA 2001) Find all strictly positive integers $x, r, p, n$ such that $p$ is prime, $n, r>1$ and $x^{r}-1=p^{n}$.
If the hypotheses of Zsigmondy's theorem are satisfied, there exists a prime factor of $x^{r}-1$ that does not divide $x-1$. Since $x-1$ divides $x^{r}-1$, this implies that $x^{r}-1$ has at least two prime factors and thus cannot be a power of a prime number.

Therefore, we need to handle the following two cases:

(i) $(x, r)=(2,6)$ (which does not work),

(ii) $r=2$ and $x+1$ is a power of 2. In this case $(x-1)(x+1)=p^{n}$, which easily gives $p=2$ and $x=3$.