Let $d = \gcd(x, y)$. Then there exist $a, b \in \mathbb{Z}$ such that $x = da$, $y = db$, $(a, b) = 1$. By substitution to the given equation we get:
$$\frac{d a (d b)^3}{d a + d b} = \frac{d^3 a b^3}{a + b} = p. \qquad (1)$$
From $(a, b) = 1$, we get $(a, a + b) = 1$ and $(b^3, a + b) = 1$, giving from relation (1) that $a + b \mid d^3$. We write
$$\frac{d^3}{a + b} = k, \qquad (2)$$
where $k$ is a positive integer. Then (1) becomes: $k a b^3 = p$, and hence $b^3 \mid p$. Therefore $b = 1$ and $k a = p$. Hence we have the following cases:
(i) If $k = p, a = 1$, then (2) becomes $\frac{d^3}{2} = p \Rightarrow 2p = d^3 \Rightarrow 2 \mid d$. Hence $8 \mid d^3$ and $8 \mid 2p$, absurd.
(ii) If $k = 1, a = p$. Then (2) becomes:
$$d^3 = p + 1 \Rightarrow d^3 - 1 = p \Rightarrow (d - 1)(d^2 + d + 1) = p.$$
Therefore, we get $d - 1 = 1$, $d^2 + d + 1 = p$ (since $d^2 + d + 1 > d - 1$),
$$\Leftrightarrow d = 2, p = 7, \text{ and so: } (x, y, p) = (14, 2, 7).$$