Since $n\geq 4$ is even, we can let $n=2k$. Then $2k\geq 4$ or $k\geq 2$ which can be substituted where we have $2^{2k}-1 = (2^k)^2 - 1^2 = (2^k+1)(2^k-1)$. Since $k \geq 2$, we have that $2^k\geq 4$.
This is where things get derailed. Where do I go from here?
If $k>1$ then $$ 4^k-1=\underbrace{(2^k+1)}_{>1}\underbrace{(2^k-1)}_{>1}. $$ If $k=1$ then $4^k-1$ is prime.
You are able to write a number as product of two numbers none of them equals to $\pm 1$.
This says it is not a prime.
Just to give a different approach, if $n=2m\ge4$, then $2^n-1\gt3$, but $2^{2m}-1\equiv(-1)^{2m}-1\equiv0$ mod $3$, so $3\mid2^n-1$.
