Huh, I never realized this was this simple. Fleshing out the general case the post mentions.

Suppose p=x^n-1 is prime.

Note that:

  x^n - 1  = (x^n + x^(n-1) +... + x^2 + x) 
            -(      x^(n-1) +... + x^2 + x +1)

  =>x^n-1  = (x^(n-1) + x^(n-2) + ... + x + 1)(x-1)

So if p=x^n-1 is prime, one of the factors must be equal to 1. If the first factor is 1, this implies x=0 and so the second factor is -1. Therefore the second factor must be 1, and x=2.

Therefore if x^n-1 is prime, x must be 2. Damn that's cool. At first I thought any number 2^n-1 would be prime, but then I realized that obviously the left hand factor can have subfactors. Which is to say I remembered 15 exists.