I was told that the proof of the Feit-Thompson theorem, which asserts that all non-abelian finite simple groups must have even order, can be greatly simplified if the following seemingly innocent number theoretic conjecture can be proved:

Conjecture: For all primes $p,q$, we have $p^q - 1 \nmid q^p - 1$.

If we suppose that there exists a counterexample, say a pair of primes $p,q$ such that $p^q - 1 | q^p - 1$, then $q^p \equiv 1 \pmod{p^q - 1}$ implies that $p | \phi(p^q - 1)$ (since by looking at the size, one can conclude that $q \not\equiv 1 \pmod{p^q - 1}$). This seems like an implausible situation by itself.

So my question is, does there exist any known examples of a prime $p$ and a positive integer $n$ (not necessarily prime) such that $\phi(p^n - 1)/p$ is an integer?

Edit: It seems that I neglected to consider the $p = 2$ case, where then the condition is trivial since $2^n - 1$ is always odd, so $\phi(2^n - 1)$ is always even. Are counterexamples easy to construct for odd $p$?