> Now I'm wondering if Fermat's Last Theorem has a bitwise analogue.

It hasn’t in this sense. The ^n operator flips the ‘one’ bits in n in its argument, so for any n, we have

  a = a^n^n

So, for all a and b,

  a^n + b^n = ((a^n + b^n)^n)^n

that makes

  c = (a^n + b^n)^n

a solution to

  a^n + b^n = c^n

and that means we can find a c for every n, a and b.