I think a more interesting problem to consider after solving n^m = m^n is solving n^m = m^n + 1 in the natural numbers.

If we exclude 0, the only solutions are 2^1 = 1^2 + 1 and 3^2 = 2^3 + 1

That is correct. Why are there no others?

Because Catalan’s conjecture.

> Catalan's conjecture was proven by Preda Mihăilescu in April 2002.

Cool, so the conjecture's a theorem.

https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu

Which has an extremely complicated proof :-(

I wouldn’t call it extremely complicated; it is much simpler than the proof of FLT.

Yes, but there is a much simpler proof.

Something related to prime factorization maybe?

Do tell.

n^n * n^(m - n) = n ^ m = m ^ n, so n^n divides m^n, so n divides m.