Thinking about $p^n = x^m + y^m$ where $p | m$ Let $p,m,n,x,y$ be positive integers.  Let $p$ be prime and $m$ be odd.
if $p \ge 5$ and $m \ge 3$ and $p | m$, are there any solution for:
$p^n = x^m + y^m$
I'm not very clear how to proceed on this type of question.  Here's what I have:


*

*$p | m$ so $p \le m$ and $m \ge 5$

*$p > 1$ so $x^m + y^m \ge 9$ and if we assume $x \ge y$, then $x \ge 2$ 

*$n > 2$ since $\dfrac{x^m + y^m}{x+y} \ge \dfrac{x^5 + y^5}{x + y} > \dfrac{x^5}{2x} = \dfrac{x^4}{2} \ge x^3 > 2x \ge x + y $
I'm not sure on the next step.
Does anyone have any suggestions or corrections?

Edit:  I am updating the question.  There may very well be solutions.  I would be very interested in finding any solution.
 A: There is no solution.
Proof:
First, let's show that $p^n=x^m+y^m$ (1) has a solution if and only if $p^n=x^p+y^p$ (2) has. Indeed, if (1) has a solution, then, write $x^m=(x^{m/p})^p$ and the same with $y$ to see that (2) has a solution. If (2) has a solution, then (1) has: just take $m=p$.  
Note that $x+y$ divides $x^p+y^p$. Then $x+y$ is a power of $p$. Say $y=p^r-x$, $r\ge 1$. Also, it is easy to see that we can suppose WLOG that $p$ does not divide $x$.
Now
$$p^n=x^p+(p^r-x)^p=\sum_{j=1}^p\binom pj p^{jr}(-x)^{p-j}\equiv p^{r+1}x^{p-1}\pmod{p^{r+2}}$$
and therefore, $n=r+1$.
So,
$$p^{r+1}=x^p + (p^r-x)^p\ge2(p^r/2)^p=p^{pr}/2^{p-1}>p^{pr-p+1}>p^{3r-2}$$
which implies $r=1$.
$$p^2=x^p+(p-x)^p\stackrel{*}{\ge} 2(p/2)^p=\frac12p^2\left(\frac p2\right)^{p-2}$$
*: Consider the function $f(t)=t^k+(c-t)^k$ for $c>0$, $k>1$ and $t\in[0,c]$. Its derivative is
$$f'(t)=kt^{k-1}-k(c-t)^{k-1}$$
This is $0$ when $t^{k-1}=(c-t)^{k-1}$, that is, when $t=c-t$ or $t=c/2$. Note also that $f'(0)=-kc^{k-1}<0$ and $f'(c)=kc^{k-1}>0$, and this implies that $f$ meets its minimum at $t=c/2$.
