Things I've gotten so far:
Proven that $2^p-1\equiv \pmod 3$ (for some odd prime)
$n$ is in the form $n =( 2^p-1)(2^{p - 1})$ where $(2^p - 1)$ is prime.$ n > 6$, so $p > 2$.
So I have the congruence $2^p-1 \equiv 1 \pmod 3$,
and I think I have to prove that $2^p -1 \equiv 4 \pmod 6$ or $-2 \pmod 6$,
but I'm not sure how to accomplish that step.
On
Observe that $$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+\cdots+x^a+1).$$ Hence a number of this form could be prime if and only if $x=2$ and $a=1$ and $b$ is a prime. Here it has shown that an even perfect number is of the form $2^{p-1}(2^p-1),$ where $2^p-1$ is a prime. Hence we only need to consider about prime powers!!
Consider $p=2$ separately (which gives you the prefect number $6$) and note that all the other primes are odd. Let $p=2m+1.$ Binomial theorem gives us,
$$2^{2m+1}=2(3+1)^m=2(3^m+\cdots+3m+1)=6.3^{m-1}+\cdots+6m+2.$$ Now $n=2^{2m}(2^{2m+1}-1)=2^{2m}(6.3^{m-1}+\cdots+6m+1).$
Hence $n \equiv 2^{2m} \pmod 6$ and you can easily prove that $2^{2m} \equiv 4 \pmod 6$ by mathematical induction.
Your formula for $n$ is wrong.
It should read $$n=(2^p-1)(2^{p-1})$$.
For example $n=28=(2^3-1)\cdot2^{(3-1)}=7\cdot4=28$
$2^{2k+1}-1\equiv 1\pmod 3$ (from Fermat's Little Theorem), and the power of $2$ forces this to be $4\pmod 6$ (from the Chinese Remainder Theorem).