How to prove indirectly that if $42^n - 1$ is prime then n is odd? I'm struggling to prove the following statement:

If $42^n - 1$ is prime, then $n$ must be odd.

I'm trying to prove this indirectly, via the equivalent contrapositive statement, i.e. that if $n$ is even, then $42^n - 1$ is not prime.
By definition, for every even number $n$ there exists an integer $k$ with $n = 2k$. We substitute and get
$$42^n - 1 = 42^{2k} - 1 = (42^2)^k - 1.$$
Now, how do I prove that $(42^2)^k - 1$ isn't a prime number? Is this even the right way to approach this proof? 

Note that 
$$42^{2k}-1=(42^k)^2-1=(42^k-1)(42^k+1)$$
where $1\lt 42^k-1\lt 42^k+1$.
