Is $(3^p-1)/2$ always squarefree?

This is even unknown for the usual Mersenne numbers $M_p=2^p-1$. All such numbers, where the factorisation is known, are squarefree. However, Guy (1994) believes that there are $M_p$ which are not squarefree. As far as I know, a similar conjecture holds for $N_p=3^p-1$ and $N_p/2$.

If $q^2$ divides $3^p-1$ with an odd prime $p$ , then $q$ must be a Wieferich-prime to base $3$. The only known Wieferich-primes to base $3$ are $11$ and $1006003$ (https://oeis.org/A014127). There is no other example below about $10^{15}$.

The example $11^2|3^5-1$ was ruled out by your condition $p>5$. There is no prime $p>5$ with $11^2|3^p-1$ because the order of $3$ modulo $11^2$ is $5$.

The order of $3$ modulo $1006003^2$ is $1006002$. Since this is not prime, $3^p-1$ cannot be divisible by $1006003^2$.

So, $3^p-1$ is almost certain squarefree for $p>5$.