Finding a prime number $p$ and $x, y, z\in \mathbb N$ such that $x^p+y^p=p^z$ I'm interested in Fermat's Last Theorem (Wiles theorem?). Then, I made the following similar question:
Find a prime number $p$ and natural numbers $x, y, z$ such that $x^p+y^p=p^z$.
I got that the followings are sufficient : for any non-negative integer $d$,$$(p,x,y,z)=(3,3^d,2\cdot3^d,2+3d),(3,2\cdot3^d,3^d,2+3d),(2,2^d,2^d,1+2d).$$ 
However, I'm not sure that these are the only solutions I want. Then, here is my question.
Question: Could you show me how to solve the above question?
 A: If $z=0$ then $x=0,y=1$ or $x=1,y=0$.
If $z \ge 1$ then $x,y \ge 1$. Let $x=p^nx_1,y=p^my_1$ with $x_1,y_1,m,n \in \mathbb{N}$ and $\gcd (x_1,p)=1, \gcd (y_1,p)=1$. The equation is equivalent to $$p^{np}x_1^p+p^{mp}y_1^p=p^{z}$$
Without loss of generality, asume that $z \ge mp \ge np$ then $x_1^p+p^{(m-n)p}y_1^p=p^{z-np}$. If $m>n$ then $p \nmid LHS$ but $p \nmid RHS$, a contradiction. Thus, $m=n$.
The equation is equivalent to $x_1^p+y_1^p=p^l \; (l \ge 1)$ with $p \nmid x_1,p \nmid y_1$ and $p|x_1+y_1$. Therefore by Lifting The Exponent Lemma we have $$v_p(x_1^p+y_1^p)=v_p(x_1+y_1)+v_p(p)=v_p(x_1+y_1)+1=l$$
Thus, $x+y=p^{l-1}$ then $l \ge 2$. It follows that $p^{l-1} \ge p$ or $x_1+y_1 \ge \dfrac{x_1^p+y_1^p}{x_1+y_1} \qquad (1)$.
Case 1. If $x_1=y_1=1$ then $x=y=p^n$. Therefore $2 \cdot p^{np}= p^z$. Thus, $p=2$ and $2n+1=z$.
Case 2. If $x_1=1$ and $y_1 \ge 2$. Hence, from $(1)$ we obtain $y_1^2+2y_1 \ge y_1^p$ or $y_1+2 \ge y_1^{p-1}$. The inequality holds if and only if $y_1=2,p=3$. Thus, $x=3^n,y=2 \cdot 3^n, z= 2+3n$.
Case 3. If $y_1=1$ and $x_1 \ge 2$. Similarly, we obtain $p=3,y=3^n,x=2 \cdot 3^n, z=2+3n$.
Case 4. If $x_1,y_1 \ge 2$. From $(1)$ we obtain $p=2$. Hence $x_1^2+y_1^2=2^l$. Since $x_1,y_1 \ge 2$ then $l \ge 2$. It follows that $4|x_1^2+y_1^2$, a contradiction since $2 \nmid x_1, 2 \nmid y_1$.
The answer is $\boxed{ (p,x,y,z,)=(3,3^d,2 \cdot 3^d,2+3d),(3,2 \cdot 3^d,3^d,2+3d),(2,2^d,2^d,1+2d)}$ for $d \in \mathbb{N}$.
A: This is a consequence of the (elementary) Zsigmondy's theorem.
Because $\gcd(x,y)$ is necessarily a power of $p$, we may suppose they're coprime.
If $n$ has at least 1 odd prime divisor, then by Zsigmondy's theorem $x^n+y^n$ has at least $2$ distinct prime divisors unless $\{x,y\}=\{1,2\}$ and $n=3$. Thus:

The only solutions to $x^n+y^n=p^z$ with $n$ not a power of $2$ are
  $$p=3,n=3,\{x,y\}=\{3^k,2\cdot3^k\},z=\ldots$$

For $n=2$ there are nontrivial solutions iff $p\equiv1\pmod4$ or $p=2$ (Fermat's Christmas Theorem) for which there is no general closed form. For $p=2$ we can note that odd squares are $1\pmod4$, so if $x,y$ are coprime, $4\nmid x^2+y^2$ and $z=1$:

The only solutions to $x^2+y^2=2^z$ are
  $$x=y=2^k,z=k+1$$

Needless to say, the general case where $n$ is a power of $2$ is related to Fermat primes and remains unsolved.
