What is known about $x^m + y^m = z^n$ over $\mathbb{N}$ when $m,n \geq 2$ and $m \neq n$? So Fermat's Last Theorem resolves the question of positive integer solutions to $x^m + y^m = z^n$ when $m = n \geq 3$. But what about if $m \neq n$ and $m,n \geq 2$? Is anything general known about when there are positive integer solutions? Clearly if $gcd(m,n) \geq 3$ then there are no solutions by Fermat's Last Theorem. So let's assume $gcd(m,n) \leq 2$.
 A: Suppose for example that $m$ and $n$ are relatively prime. Let $x=y=2^s$ and $z=2^t$. Then our equation becomes $2^{1+sm}=2^{tn}$. 
There are infinitely many pairs of positive integers $(s,t)$ such that $1+sm=tn$, so the equation $x^m+y^m=z^n$ has infinitely many positive (but uninteresting!) solutions. 
A: Equation of the form
$$x^p + y^q = z^r\quad\text{ where }\quad x, y, z \in \mathbb{Z}\;\;\text{ and }\;\; p, q, r \ge 2,\; \max(p,q,r) > 2$$
is called super-Fermat equation. A solution of it is called trivial if $xyz = 0$ or $\pm 1$ and called primitive if $\gcd(x,y,z) = 1$.
For the equation at hand
$$x^m + y^m = z^n \quad\text{ where }\quad m, n \ge 2, m \ne n, \gcd(m,n) \le 2$$
Following is a summary of what I know:


*

*$x^3 + y^3 = z^2$
There are infinitely many primitive solutions falling into 3 infinite families.
One infinite family is given by the parametrization:
$$\begin{cases}
x &= u(u^3 − 8v^3),\\
y &= 4v(u^3 + v^3),\\ 
z &= u^6 + 20u^3v^3 − 8v^6
\end{cases}$$
subject to the constraint $3 \not| u + v$, $2 \not| u$ and $\gcd(u,v) = 1$.

*$x^m + y^m = z^2$ for $m > 3$.
There is no non-trivial primitive solution. See the paper Winding quotients and some variants of Fermat’s Last Theorem 
by Darmon and Merel for a proof.

*$x^2 + y^2 = z^n$ for $n \ge 3$.
There are infinitely many primitive solutions. In particular, given any two positive integers $s,t$ one even, one odd, co-prime with each other. If we define  $x,y,z$ by
$$x + i y = (s + it)^n \quad\text{ AND }\quad z = (s^2+t^2)$$
then $x,y,z$ will be a primitive solution.

*Otherwise, $m \ne n$ and  $m, n \ge 3 \implies 1/m + 1/m + 1/n < 1$.
It is known that if we assume abc conjecture, then the total number of non trivial primitive solutions for
$$x^p \pm y^q \pm z^r = 0\quad\text{ with }\quad 1/p + 1/q + 1/r < 1\tag{*1}$$
is finite, even allowing $p, q, r$ to vary. Here, if $x = \pm 1$ (respectively $y = \pm 1$ or $z = \pm 1$), we identify solutions having the same value of $x^p$ (respectively $y^q$ or $z^r$)
As of this moment, there are essentially only 10 known solutions for $(*1)$:
$$\begin{array}{rcll}
1^p + 2^3 &=& (\pm 3)^2  &\text{ for $p \ge 7$, and $(-1)^p$ when $p$ is even}\\
(\pm 3)^4 + (-2)^5 &=& (\pm 7)^2\\
2^9 + (-7)^3 &=& (\pm 13)^2\\
2^7 + 17^3 &=& (\pm 71)^2\\
3^5 + (\pm 11)^4 &=& (\pm 122)^2\\
15613^3 - (\pm33)^8 &=& (\pm1549034)^2\\
65^7 + (-1414)^3 &=& (\pm 2213459)^2\\
113^7 + (-9262)^3 &=& (\pm 15312283)^2\\
17^7 + 76271^3 &=& (\pm 21063928)^2\\
(\pm 43)^8 + 96222^3 &=& (\pm 30042907)^2
\end{array}$$
Since all these known solutions has at least one of $p, q, r$ equal to $2$, this means
there are no known non-trivial primitive solutions when both $m, n \ge 3$.
Most of the material in this post is extracted from Chapter 14, The super-Fermat equation of the book Number Theory, Vol II: Analytic and Modern Tools by H. Cohen. If you want more details, please consult the book directly.
A: I attempt to solve the equation $x^3+y^3=z^n, x,y,z, n\in N \textrm{  and }n>3.$
Noting that $(1,1,2,1)$ is a solution and multiplying it by $2^{3k}$, we have
$$1^3+1^3=2^1 \Rightarrow \left(2^k\right)^{3 }+\left(2^k\right)^3=2^{3 k+1}$$
Therefore there are infinitely many solutions
$$(2^k,2^k, 2,3 k+1),$$
where $k\in N.$

Also $(1,2,3,2)$ is a solution and multiplying it by $3^{3k}$, we have
$$1^3+2^3=3^2 \Rightarrow (3^k)^{3 }+\left(2 \cdot 3^k\right)^3=3^{3 k+2}$$
Therefore there are infinitely many solutions
$$(3^k,2(3^k), 3,3 k+2) \textrm{ or } (2(3^k) ,3^k, 3,3 k+2), $$ where $k\in N.$
