If $n$ is an even integer greater than $2$, then $2^n - 1$ is not a prime. Fairly new to Discrete Mathematics and I'm stumped on this one. So we're asked to prove:

If $n$ is an even integer greater than 2, then $2^n - 1$ is not a prime.

What I can come up with is that since $n > 2$, we know that $n$ is not prime since the only even $n$ happens to be $2$. We can write $n = 2k$ and so we rewrite $$2^n - 1 = 2^{2k}-1 = (2^k)^2 - 1 = (2^k-1)(2^k+1)$$
Up to here, am I even remotely correct? I'm not sure what else to say to take it from here to fully prove this. I also apologize for how I worded it, as I'm still trying to understand how to explain my proofs.

You are doing fine.  Now that you have shown a factorization of $2^n-1$ the only thing that can go wrong is that one of the factors is $1$.  So if $n \gt 2, \ldots$  
A more general statement is that if $n$ is composite, $2^n-1$ is never prime.  The reasoning is the same.  If $n=ab$ with $a,b \gt 1$ then $2^n-1$ is divisible by $2^a-1$ and $2^b-1$.  You can just do the division or search this site for the proof.  Your problem is the $a=2$ case of this.
