Integer is powerful if all the exponents in its factorization are at least $2$.

Every powerful integer can be written in the form $a^2 b^3$.

For odd $k$, define $F(k)=(2^{2k}-1)=(2^k-1)(2^k+1)$.

This paper asks are there only finitely integers for which $F(k)$ is powerful.

It is natural to ask what properties the smallest solution to $F(K)=\textit{powerful}$ has.

$2^k+1$ is always divisible by $3$ for odd $n$ and it is divisible by $3^2$ iff $k$ is divisible by $3$. This shows $3 \mid K$, so there are no solutions with $K$ prime.

Observation 1: if odd $n \mid k$, then $2^n-1 \mid 2^k -1$ and $2^n+1 \mid 2^k+1$.

Observation 2: if odd prime $q^2 \mid 2^{2k}-1$ and $q$ is not Wieferich prime, then $q \mid k$ since the multiplicative order of $2$ modulo $q^2$ is divisible by $q$.

Conjecture 1: Assume $q$ is Wieferich prime and $q^2 \mid 2^k-1$. Then $d \mid k$ where $d$ is divisor of $q-1, d > \log_2{q}$.

We believe the observations and the conjecture imply infinitely many non-Wieferich primes.

So starting from $3 \mid K$ we are removing primes with exponents one and continue $2^3-1 \mid K, 2^7-1 \mid K, 2^7+1 \mid K, 2^{127}-1 \mid K, 2^{127}+1\mid K$.

This constraints are very strict since $2^{2^{127}-1}-1$ is extremely large.

Q1 Are these constraints true?

Q2 Can we get more constraints, hopefully showing no solutions?