It is not known if mersenne numbers with prime exponent are square free. It is an open problem in number theory. Some limitations on the divisors are discussed in: https://mathoverflow.net/questions/149511/squarefree-parts-of-mersenne-numbers
Do we know if a mersenne numbers with prime exponent are cube free?
Not a full answer, but a proof that $2^p\equiv 1\mod q^2$ with primes $p,q$ implies that $q$ is a Wieferich prime.
Define $m=ord_2(q^2)$ , in other words, $m$ is the smallest positive integer with $$2^m\equiv 1\mod q^2$$
Because of $$2^p\equiv 1\mod q^2$$ the order must be $1$ or $p$. $1$ can be ruled out because of $2\ne 1\mod q^2$ , hence the order must be $p$.
Euler's theorem states $$2^{q(q-1)}\equiv1\mod q^2$$ hence $p|q(q-1)$.
$p|q$ would imply $p=q$ , but $2^q\equiv 2\ne 1\mod q$ , hence $2^q\ne 1\mod q^2$. Hence $p$ must divide $q-1$ implying $$2^{q-1}\equiv 1\mod q^2$$
Hence $q$ is a Wieferich-prime.