Let $M_n$ denote the Mersenne numbers $M_n=2^n-1$.

As $n$ varies, must $M_n$ be divisible by arbitrary large prime $p$ with exponent one, i.e. $p \mid M_n, p^2 \nmid M_n$?

In other words, must the exponential diophantine equation $2^n-1 = A x^2 y^3$ for constant $A$ has only finitely many solutions $(n,x,y)$ and $n >1$?

Related question is this.

Also related question that might show there are infinitely many non-Wieferich primes is here.

Looking for unconditional results, abc easily implies it.

**Added**

The paper Remarks on Exponential Congruences and Powerful Numbers P. RIBENBOIM on p7.

(M') There exist infinitely many Mersenne numbers which are not powerful.

(M') implies B_2 implies infinitely many non-Wieferich primes.

One easy way to construct infinitely many non-powerful Mersenne numbers is to observe that 3 divides $M_{6n+2}$ with exponent one, so $M_{6n+2}$ is not powerful.

Doesn't this approach give infinitely many non-Wieferich primes?