Solve $x^p + y^p = p^z$ when $p$ is prime Find the solutions in positive integers of $x^p + y^p = p^z$, where $p$ is a prime number.
Particular case $p=2$:


*

*For $z=0$ there are no solutions.

*For $z=1$ the only solution is $x=y=1$.

*For $z\ge2$ the solutions are $(x,y,z)$ $(2^k,2^k,2k+1)$.
Now the general case $p \gt 2$:
$p$ is odd 
so $(x + y)(x^{p-1} -x^{p-2}y +...-xy^{p-2} + y^{p-1})= p^z$
being $p$ a prime we have $x + y= p^m$ and $\sum_{k=0}^{p-1}(-1)^kx^{(p-1)-k}y^k = p^n$ 
with $m + n = z$
Now i'm stuck and nothing come to my mind... also a substitution like $y= p^m -x$ doesn't seem the right path...thanks to those who'll help.

If you use the well known lemme

Lemma 1: Lifting The Exponent Lemma (LTE) :Let 
  $x$ and $y$ be (not necessary positive) integers, let $n$ be a positive integer, and let $p$ be an odd prime such that $p\mid x−y$ and none of $x$ and $y$ is divisible by $p$ (i.e.,$p\not \mid x$ and $p\not \mid y$). then:
  $$v_p(x^n-y^n)=v_p(x-y)+v_p(n) $$
Lemma 2: For very positive integers $x\geq 2$ and $n\geq 2$ we have $x^n\geq nx$ with equality if and only if $x=n=2$ or $x=1$

Proof (Lemma 2) using Bernoulli's inequiality we have:
$$x^n=x(1+(x-1))^{n-1}\geq x(1+(x-1)(n-1))\geq x(1+(n-1))=xn $$
Let $x,y,z$ be positive integers and $p$ an odd prime number such that:
$$x^p+y^p=p^z\ \ \ \gcd(x,y)=1$$
Because $z$ is a positive integer and $(x+y)$ divides $x^p+y^p=p^z$ then there exists an integer $m$ such that $x+y=p^m$, using the lemma we have:
$$z=v_p(x^p+y^p)=v_p(x+y)+v_p(p)=v_p(x+y)+1=m+1$$
so we have : $x^p+y^p=p^{m+1}$ and $x+y=p^m$ so two cases :


*

*If $x\geq 2$ and $y\geq 2$ (and because $p\geq 2$) we can apply lemma $2$ so $x^p\geq px$ and $y^p\geq py$ so that $x^p+y^p\geq p(x+y)$ , but we have equality in this inequality then $p=x=y=2$ which is impossible because of $\gcd(x,y)=1$

*If $x=1$ or $y=1$ we can suppose WLOG that $x=1$ (symmetry of the problem on $x,y$) and therefore $y=p^m-1$ and hence:
$$p^{m+1}-1=(p^m-1)^p$$
but we know that $p^{m+1}>p^{m+1}-1=(p^m-1)^p>p^{(m-1)p}$ hence $m+1>(m-1)p$ so $m\leq 1$ so we must have $m=1$  ($m=0$ is impossible) and then $p+1=(p-1)^{(p-1)}$ but this gives us $p=3$ (otherwise $p-1> 2$ and $(p-1)^{p-1}-(p+1)\geq (p-1)^2-(p+1)=p(p-3)>0$)


Now we found all primitive solutions for odd primes $(2,1,2),(1,2,2)$ with $p=3$ which gives the the general solution:
$$(2.3^{k},1.3^{k},3k+2),(3^k,2.3^k,3k+2)$$
