You have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.
There are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.
Version 1: If $n$ is prime, then for any integer $a$, we have $a^n \equiv a \pmod n$. This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.
Version 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \equiv 1 \pmod n$.
In your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$.
However, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows.
Since $p=2^n-1$, certainly $p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$. That implies that the order of $2$ in the field $\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$.
The order of any element of the field $\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$.
Your proof of the first statement is incomplete, since you only deal with numbers which can be written as the product of two prime numbers. The idea is fine, though. It is easy to prove by induction that the product of any number of numbers congruent to $1$ modulo $4$ is again congruent to $1$ modulo $4$.
The other proof is fine.
Best Answer
squares are $0,1,4 \pmod 5$ so $3m^2 \equiv 0,3,2 \pmod 5,$ next $1+3m^2 \equiv 1,4,3 \pmod 5$ Finally $$ -(1+3m^2) \equiv 4,1,2 \pmod 5 \; , \; \; $$ $$ 2 n^2 \equiv 0,2,3 \pmod 5 $$
The overlap of these two lists, $4,1,2$ and $0,2,3$ is the single possibility $2.$ That is, we need $2n^2 \equiv 2 \pmod 5$ and $n^2 \equiv 1 \pmod 5$