Today, I am posting an approach that seems too elementary to validate. The approach is developed by me while reading the 1967 work of L. Warren and H. Bray. In their article, "On the square-freeness of Fermat and Mersenne numbers", they have established interesting congruence relations for the primes whose square divides these Fermat and Mersenne numbers.
However, in the first place, I felt, I must give my take on the problem of square-freeness of Mersenne Numbers. Here, in this script, I've discussed my approach to the problem.
To prove: Every number $M_p = 2^p -1$ is square-free for all primes $p$.
Proof:
We assume that $M_p$ is not square-free, to proceed with the method of contradiction. This implies that,
$$M_p = (8c+1) (4k \pm 1)$$
Here, $(8c+1)$ comes from the fact, that odd-square is of form $8c+1$. And the other factor, $4k \pm 1$ is simply assumed because every odd number can be represented in this form.
Case 1: $2^p -1 = (8c+1)(4k+1)$
$2^p -1 = 32ck + 8c + 4k + 1$
$2^p = 32ck + 8c + 4k + 2$
$2^{p-1} = 16ck + 4c + 2k + 1$
$2^{p-1} = 2q + 1$
Here, $q = 8ck + 2c + k$. However, no even number is never equivalent to an odd number. Hence, a contradiction.
Case 2: $2^p -1 = (8c+1)(4k-1)$
$2^p-1 = 32ck - 8c + 4k -1$
$2^p = 32ck - 8c + 4k$
$2^{p-2} = 8ck - 2c + k$
Now, if we compute, Euler totient function $\phi$ of above equation, we've,
$2^{p-3} = \phi(8ck- 2c + k)$.
But, $2^{p-2} = 2^{p-3}*2 = 8ck -2c + k$. Hence, $2^{p-3} = 4ck - c + \frac{k}{2}$. Now, notice that left hand side of last equation is an integer for all values of $p$ which makes $k = 2r$ for some some integer $r$. Ultimately, we have, $2^{p-3} = 8cr - c + r$.
Now, we let $m = 8cr - c + r$.
Using the last equation and the equation obtained by computing the Euler totient function, we have,
$$\phi(2m) = m$$.
For above to be true, $gcd(2,m) = 1$, which implies that $m$ is an odd number. Recalling, $m = 8cr - c + r = 2^{p-3}$. Hence, a contradiction.
This completes the proof.
The approach seems to be too elementary to me. I will be thankful to the Stack community if they can review this approach and give their valuable insights on the same.