One case where this is true is $z=1$:
$$(1)^{m-1} = \frac{1^m + 1^m}{1+1}$$
Another case is where $x=y$:
$$(x)^{m-1} = \frac{x^m + x^m}{x+x}$$
Assuming that $x,y,z,m$ are positive integers with $z > 1, m \ge 3$, gcd$(x,y)=1$, is it ever the case that:
$$z^{m-1} = \frac{x^m + y^m}{x+y}$$
How would one go about solving for this situation or proving that there are no solutions?
If $z > 1$, then there exists positive integers $a,b$ such that $z=a+b$
$$(a+b)^{m-1} = a^{m-1} + {{m-1} \choose 2}a^{m-2}b + \dots + b^{m-1} = x^{m-1} - x^{m-2}y + \dots + y^{m-1}$$
I can see that $a+b < x+y$. I'm trying to see if there is a next logical step to take to make progress.
Is there a standard solution for this type of question? Does anyone have any tips on how to proceed on analyzing this type of problem?
Edit: Updated the question based on a comment.