Arising from this recent question, and in particular the answer by Gerry Myerson, it occurs to me that, if $m$ and $n$ are coprime integers, non-trivial solutions can be found to any Diophantine equation of this form (or with more powers of $m$ on the left hand side):
$$a^m + b^m = c^n$$
The method is to take any integers d and e, find $f = d^m + e^m$, then find an integer $r$ such that $r$ is congruent to $0 \mod m$ and to $-1 \mod n$. The Chinese Remainder Theorem tells us that such an integer can be found. Then:
$$(d^m)(f^r) + (e^m)(f^r) = f(f^r)$$
has powers of m on the left and a power of n on the right.
Is this correct, or am I missing something? (I realise that to find solutions with no common divisor is much harder or perhaps impossible.)