$2^n + 3^n = x^p$ has no solutions over the natural numbers A few weeks ago, I was asked to prove that $2^n + 3^n = x^2$ has no solutions over the positive integers. My proof was:
$2^n + 3^n \equiv (-1)^n \equiv \pm 1 \mod{3}\\\text{However, quadratic residue mod 3 is always 0 or 1. Therefore:}\\n = 2m\\\text{Now, working mod 5:}\\2^{2m} + 3^{2m} \equiv 4^m + 9^m \equiv 2 \times(-1)^m \equiv \pm 2 \mod{5}\\\text{However, quadratic residue mod 5 is }0, \pm 1\\\text{As such, } 2^n + 3^n \text{ is never a perfect square.}$
Naturally, I questioned if the $2^n+ 3^n$ is ever a perfect cube, which a friend and I managed to prove was false, by working mod 7 and mod 9, and using Fermat's Last Theorem. This led us to consider the equation $2^n + 3^n = x^p,\,(n,x,p)\in \mathbb{N}^3,\,p\geq2$ and conjectured that there are no solutions. Obviously, if $p$ is composite and there are no solutions for any one of its factors then there are no solutions for p (e.g. $p = 4$ has no solutions because $x^4 = (x^2)^2$, so it is only necessary to consider the cases where $p$ is prime.
$p = 5$ was next, and we managed to partially prove it by working mod 11 and 25, and again, Fermat's Last Theorem, leaving only the case $n\equiv 6 \mod{10}$.
So my question is: is $2^n + 3^n$ ever a perfect power, and if not, how can this statement be proved? Alternatively, if this statement is false, are there any strategies other than using a computer to find a counterexample?
 A: 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.
