Proof involving prime numbers

I need to prove that if either of the numbers $\displaystyle 2^n-1$ and $\displaystyle 2^n+1$ is prime for $\displaystyle n>2$, then the other one is not.

Attempt: If we name them $\displaystyle a=2^n-1$ and $\displaystyle b=2^n+1$, we realize that $\displaystyle b=a+2$. And the the smallet prime for the first number is when n=3, so $\displaystyle a=2^3-1=7$. For the other number the first prime occurs at n=4 and it's 17.

So, how should I prove this question? If we suppose $\displaystyle 2^n-1$ is prime, it will have only the two divisors $\displaystyle \{ 1, 2^n-1 \}$. Then what does that say about $\displaystyle 2^n+1$?

(We can't say that the second number is not prime because it is two digits larger than the first one, since many prime numbers are distrubuted at intervals of 2. And there is a chance it can be prime since it will be odd).