$2^n + 3^n = x^p$ has no solutions over the natural numbers

I'll provide a (relatively) simple proof (with the hard work hidden!) tailored for this problem - a similar argument works for more general equations such as, for example, $$ 2^m \pm 3^n = x^p. $$ Suppose we have a solution to the equation $$ 2^n+3^n=x^p, $$ where $n$ and $x$ are positive integers and $p$ is prime. Then, as noted by the proposer, we may, through congruence arguments, assume that $p \geq 5$. We may also suppose that $n \geq 5$. Consider the "Frey" curves $$ E_1 \; \; : Y^2 = X (X-3^n) (X+2^n), $$ if $n$ is odd, and $$ E_2 \; \; : Y^2 = X (X+x^p) (X+2^n), $$ if $n$ is even (we consider these two cases separately to get the smallest conductor we can). Then it can be shown (by Tate's algorithm) that each curve has conductor $$ N_{E_i} = 6 \prod_{q \mid x} q, $$ where the product is over prime $q$. Since we suppose that $p \geq 5$ (which guarantees that certain Galois representations that I haven't defined are irreducible), we may appeal to Ribet's level lowering results to conclude that these Frey curves correspond to certain weight $2$ modular forms of level $6$. This is a contradiction since the space of such forms turns out to be empty.