Answer     (2, 3)

Starting by the guessing and checking, we find only $2^1 = 1^3 + 1^3$ and $3^2 = 2^3+1^3$. Now we prove there are the only solutions.

Assume there exist positive integers $(p, x, y, n)$ satisfying the above equation where $n$ is the smallest possible value and prime $p > 3$. It also obviouse that $x+y > 3$ because all the possible combination of $(x, y)$ such that $x+y\le 3$ cannot meet the requirement, Similarly, we can also find $(x-y)^2 +xy \ge 2$.

By $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = p^n$ where $p$ is a prime, and $x+y \ge 3$ and $x^2 - xy + y^2 = (x-y)^2 + xy \ge 2$. Hence both $(x+y)$ and $(x^2 - xy + y^2)$ are divisible by $p$. It follows that $(x+y)^2 - (x^2-xy+y^2)=3xy$ is also divisble by $p$. Because $3 \mid p$, one of $x$ and $y$ must be divisible by $p$. Because $p \mid x+y$, we find both $x$ and $y$ are divisible by $p$. Then $x^3 +y^3 \ge 2p^3$. So we must have $n > 3$ and $$p^{n-3} = \frac{p^n}{p^3}=\frac{x^3}{p^3} + \frac{y^3}{p^3} = \Big(\frac{x}{3}\Big)^3+\Big(\frac{y}{3}\Big)^3$$

This contracts to the assumption that $n$ is the smallest.