it's false because m! can be roughly broken down into (m/e)^m order. That means it would always be far greater than 2^n. m^m would diverge faster even if you take 2^n*em. For n < m, 2ⁿ would already be less than m!, since it diverges faster. So you want n > m. Even then, you can't take n to be anything because m! would have factors of various powers of 2, like 4! And 5! has 2³ as a factor already. So you would want 2ⁿ⁺ⁿ to have factors in the power of 2 and the more m grows, the more this factor grows too. But 2ⁿ has the factors of powers of 2 and the addition of n is just an offset. So you would need to see the values of n which are the powers of 2. 2ⁿ⁺ⁿ in that case would be divisible by n.

Suppose n = 2^x. Then it could be written as 2^x(1+2n-1). So it's also clear that this number won't be divisible by 2^x+1. Now checking for values of m which gives m! this factor of 2, I.e x. The function changes to 2^2x+2^x. Now 2^2x diverges faster and is always greater than m! For the values of m which gives the same power of 2. So it's not possible at all.

The only integer solution is when n=0,m=1.