I am wondering if there is a way to prove the following statement, which bears some resemblance to Zsigmondy's Theorem. I am not sure if the statement is true, but it seems as though it should be.
For any prime $p$, there is some positive integer $n$ such that $p^{2^n}-1$ is divisible by two distinct primes that do not divide $p^k-1$ for any $k\in\{1,2,\ldots,2^n-1\}$.
Indeed, Zsigmondy's Theorem guarantees that $p^{2^n}-1$ will be divisible by some prime that does not divide $p^k-1$ for any $k\in\{1,2,\ldots,2^n-1\}$ no matter the choice of $n$ (except, perhaps, if $n=1$ and $p$ is a Mersenne prime). One might also phrase the problem as follows:
Let $p$ be a prime. Show that there is some positive integer $n$ such that $ord_{q_1}(p)=ord_{q_2}(p)=2^n$ for distinct primes $q_1$ and $q_2$.
