Are all Mersenne Numbers with prime exponent square-free?

Definitions and mathematical statement

For a prime exponent $p$, set $M_p=2^p-1$. The question is whether every such integer is square-free: \[ \forall\text{ primes }p,q,\qquad q^2\nmid 2^p-1. \] The exponent $p$ and a possible divisor $q$ are distinct roles and need not be equal. Square-free means that every prime factor occurs with multiplicity exactly one. It allows $M_p$ to be composite. Restricting the exponent to primes is part of the problem, rather than a conclusion to be established.

For $p=2$, one obtains $M_p=3$. For odd prime $p$, any prime divisor $q$ of $M_p$ is odd and the multiplicative order of $2$ modulo $q$ is exactly $p$: the order divides $p$ and cannot be one. Thus $p\mid q-1$. This necessary divisibility condition explains why the two prime parameters cannot be treated as arbitrary divisors of a general Mersenne number. A counterexample requires the stronger congruence $2^p\equiv1\pmod{q^2}$ for an actual prime $p$. Checking square-freeness for any bounded set of exponents leaves the universal assertion beyond that bound untouched.

Short English statement

Must every Mersenne number with prime exponent have all its prime factors occurring only once?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3374 (OPG-59976), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Are all Mersenne Numbers with prime exponent square-free?. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/are_all_mersenne_numbers_with_prime_exponent_square_free.