When is a prime factor of Mersenne number Wieferich prime?

Question

Wieferich prime is a prime number $p$ such that $p^2$ divides $2^{p - 1} - 1$.

There are only two Wieferich primes known and it is an open problem if there are infinitely many non-Wieferich primes.

Mersenne numbers are $M_n=2^n-1$.

When is a prime factor of Mersenne number Wieferich prime?

Wikipedia claims

A prime divisor $p$ of $M_q$, where $q$ is prime, is a Wieferich prime if and only if $p^2$ divides $M_q$.

Wikipedia reference doesn't appear to work, but we found it in paper:

link PRIME POWER DIVISORS OF MERSENNE NUMBERS AND WIEFERICH PRIMES OF HIGHER ORDER, Ladislav Skula

Confusion is possible, but we believe that Wikipedia's claim implies infinitely many non-Wieferich primes.

Assume the set of non-Wieferich primes is finite and let $P$ their product.

Then for all primes $q$, $M_q=d u$ where $d$ is divisor of $P$ and $u$ is product of powers of Wieferich primes with exponents at least $2$.

If all exponents are $2$, this implies that the squarefree free part of $M_q$ is bounded, which is easily shown to be impossible e.g. see this queston

What is wrong with this argument?

Answer 1

First, every odd prime divides some Mersenne number. (In fact, every odd integer divides infinitely many Mersenne numbers.) So that is no restriction at all.

Second, the claim from Wikipedia is correct, for the reasons given in CHUAKS answer. We can improve the Wikipedia claim to the following equivalence: A prime $p$ is Wieferich if and only if $p^2|(2^{{\rm ord}_p(2)}-1)$. Here ${\rm ord}_p(2)$ is the order of $2$ modulo $p$, and it is a divisor of $p-1$. (Generally, it is not necessarily prime.)

Third, Wieferich primes can potentially occur to odd exponents in $M_q$, as pointed out by Alexander Kalmynin, which is why your argument breaks.

Answer 2

The first question has an answer. If $p$ is an odd prime, $p \nmid a$ and $p \mid (a^n-1)$, then $p$ is Wieferich to the base $a$ if and only if $\nu_p(a^n-1) \ge \nu_p(n)+2$ which follows from an identity :

$\nu_p(a^n-1)=\nu_p(n)+\nu_p(a^{p-1}-1),$

which is (1.1) in https://arxiv.org/abs/2112.04173.