Using Quadratic Residue to show that $2^m-1$ doesn't divide $3^n-1$ Show that if $m$ is congruent to $n$ modulo $2$ and $m>1$ then
$2^m-1$ cannot be a divisor of $3^n-1$
What I have tried:
I showed that if $p$ is a divisor of $2^m-1$ then
if $m$ and $n$ are odd then $2$ and $3$ are quadratic residue of $p$ if $p$ is congruent to $1$ or $-1$ modulo $12$ but I cannot see how to proceed.

With both $m$ and $n$ odd, since $m \gt 1$ so $2^m - 1 \gt 1$, for any prime factor $p \mid 2^m - 1$, since $p \mid 3^n - 1$ as well, then both $2$ and $3$ must be quadratic residues modulo $p$ (as you already stated). The table in the Law of quadratic reciprocity section shows $2$ is a quadratic residue mod $p$ if and only if
$$p \equiv 1, 7 \pmod{8} \tag{1}\label{eq1A}$$
and $3$ is a quadratic residue mod $p$ if and only if
$$p \equiv 1, 11 \pmod{12} \tag{2}\label{eq2A}$$
Since $11 \equiv 3 \pmod{8}$ and $13 \equiv 5 \pmod{8}$, combining the requirements in \eqref{eq2A} and \eqref{eq3A} gives
$$p \equiv 1, 23 \pmod{24} \tag{3}\label{eq3A}$$
Since $23 \equiv -1 \pmod{24}$, the product of any number of these factors will be $1$ or $-1$ mod $24$. However, for odd $m \gt 1$, you can easily show, such as by induction using $(2^2)(8) \equiv 8 \pmod{24}$, that
$$2^m - 1 \equiv 7 \pmod{24} \tag{4}\label{eq4A}$$
Thus, there must be at least one prime factor of $2^m - 1$ which is not congruent to either $1$ or $-1$ modulo $24$. This therefore means
$$2^m - 1 \not\mid 3^n - 1 \tag{5}\label{eq5A}$$
