Solving $x^p + y^p = p^z$ in positive integers $x,y,z$ and a prime $p$

The question is from Zeitz's ''The Art and Craft of Problem Solving:"

Find all positive integer solutions $x,y,z,p$, with $p$ a prime, of the equation $x^p + y^p = p^z$.

One thing I noticed is that

$$ \frac{x^p + y^p}{x + y} = \sum_{i=0}^{p-1}x^{(p-1)-i}(-y)^i \implies (x+y) |p^z \implies x + y = p^n, \text{ }n < z $$

It is also not hard to determine all solutions for $p =2$. After this, however, I am at a loss. I don't know what restrictions I can impose to try to narrow down the solution set; for example, the class of solutions

$$ p=3, x = 3^n, y = 2\cdot3^n, z = 2 + 3n $$

for $n \ge 0$ show that $x+y = p^n$ cannot be sharpened. (Let me note that these are the only other solutions I have found besides the ones for $p = 2$.)

Could anybody give me a push in the right direction on this problem?

Sorry, I can't think of a good hint to give you a push. Here is my solution. There likely is a better way of approaching it, since this is from Art and Craft.

Deal with the case $p=2$ separately. Henceforth, $p$ is an odd prime.

If $\gcd (x, y) = k > 1$, then $k^p \mid p^z$, which implies that $k \mid p^z$. Thus, we can divide out by $k$. Henceforth, assume that $\gcd(x,y) = 1$.

Since $x + y \mid x^p + y^p$, hence $ x+y = p^n$. We have

$$x^p + (p^n - x)^p = p^z.$$

With the condition that $\gcd(x,y) = 1$, we get that $ p \not \mid x$. As such, $p^{n+1}$ divides $x^p + (p^n - x)^p$ but $p^{n+2}$ doesn't. Hence,

$x^p + (p^n - x)^p = p^{n+1} $.

This becomes extremely restrictive. For $p \geq 5$, we have

$LHS \geq \frac{p^{5n}}{16} > p^{n+1}=RHS$,

Hence, the only possibility is $p=3$.