Show that if 2^m is an odd prime then m = 2^n for some nonnegative integer n

Okay, im not sure if this is what you needed but if we are saying its (2^m) + 1, we know that m={1,2} will all result in an odd prime. You can use log₂ (m) = n and you will find that for the sequence m={1,2}, the corresponding n values are n={0,1} which are nonnegative integers.

I do not know if you need to show proof but both m={1,2} and the corresponding n values show that this works.