For fixed $m$, find all positive integer solutions to $a^m+b^m = p^n$ where $p$ is prime The problem:
For fixed $m$, 
find all solutions to 
$a^m+b^m = p^n$ 
where $p$ is prime,
all variables are positive integers,
 $a \le b$,
and
$m \ge 2$.
This is a generalization 
of this question:
Solve $a^{2013}+b^{2013}=p^n$ for all positive integers a,b,n and prime numbers p
My answer showed that
if $m$ is odd
and divisible by 3
then the only solution
(aside from the trivial one of
$a=b=1, p=2, n=1$)
is
$a=1, b=2, m=3, p=3, n=2$
(i.e.,
$1+8=9$).
I am now asking about
the solutions for
other values of $m$.
Some observations:
If $a = b$,
the equation is
$2a^m = p^n$,
so $p=2$,
$a=2^k$,
and
$n = km+1$.
If $m$ is even,
then $p^n$ is the
sum of two squares,
so the results there
can be used.
In particular, if $p$
is of the form
$4k+3$,
then $n$ must be even.
For what other classes of $m$,
or even particular values of $m$,
can all solutions be found?
 A: The only positive integer solutions of $p^n=a^m+b^m$ with $p$ prime, $m\ge 3$ not a power of $2$ are $(p,n,a,b,m)=(3,2,2,1,3),(2,km+1,2^k,2^k,m),\forall k\ge 1$.
If $a=b$, then $(p,n,a,b,m)=(2,km+1,2^k,2^k,m), k\ge 1$. If $(a,b,m)=(2,1,3)$, then $(p,n,a,b,m)=(3,2,2,1,3)$. Otherwise: let $m=2^kt$ with $t\ge 2$ odd. Then by Zsigmondy's Theorem:
$$a^m+b^m=\left(a^{2^k}+b^{2^k}\right)\left(a^{m-2^k}-a^{m-2^k-1}b\pm\cdots+b^{m-2^k}\right)$$
has a prime divisor that does not divide $a^{2^k}+b^{2^k}$, so $a^m+b^m$ has at least two different prime divisors.

However, this is not so simple if $m$ is a power of $2$. Some partial results:
If $m=2$, then $a^2+b^2=p^n$ has infinitely many positive integer solutions with $p$ prime. E.g., if $n=1$, then there exists a unique solution if $p\equiv 1\pmod{4}$; no solutions if $p\equiv 3\pmod{4}$. See Fermat's theorem on sums of two squares (or Fermat's Christmas Theorem). Uniqueness isn't mentioned in the Wikipedia page, but it's easy to prove; see e.g. here (there's a small mistake; it should be $(ad+bc)^2+(ac-bd)^2$).
If $m=2, n=2$, then $a^2+b^2=p^2$ has infinitely many positive integer solutions. It's well-known (see here) that we must have $p=k(m^2+n^2)$ for some positive integers $k,m,n$ (wlog $m>n$). Clearly $k=1$. If we can find such $m,n$, then $\{a,b\}=\{2mn,m^2-n^2\}$ solves the equation. So $a^2+b^2=p^2$ is solvable iff $p\equiv 1\pmod{4}$.
If $m=4$, in $1961$ it was unknown whether there are inifinitely many primes of the form $a^4+b^4$ (see here). If $m=2^t$ with $t\ge 2$, $n$ must be odd (see here).
Also see Generalized Fermat Numbers. In particular, it is open whether there are infinitely many primes of the form $a^{2^n}+1$, $n\ge 1$, in particular if there are infinitely many primes of the form $n^2+1$ (see here).
If $\gcd(b,p)=1$, we must have $p\equiv 1\pmod{4}$, because $-1$ is a quadratic residue mod $p$ iff $p\equiv 1\pmod{4}$ (Quadratic Reciprocity).
