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?
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?