4. (Bulgaria) Determine all positive integers $x, y$, where $z$ is the greatest common divisor of $x, y$, that satisfy the equation $x+y^{2}+z^{3}$ $=x y z$.

保留了原文的换行和格式。
Let $x = zc, y = zb$, where $c, b$ are coprime integers. Then the given Diophantine equation can be transformed into $c + zb^2 + z^2 = z^2cb$. Therefore, there exists some integer $a$ such that $c = za$. Thus, we get: $a + b^2 + z = z^2ab$, i.e., $a = \frac{b^2 + z}{z^2b \cdots 1}$. If $z = 1$, then $a = \frac{b^2 + 1}{b^2 - 1} = b + 1 + \frac{2}{b - 1}$, from which we can see that $b = 2$ or $b = 3$. Therefore, $(x, y) = (5, 2)$ or $(x, y) = (5, 3)$. If $z = 2$, then $16a = \frac{16b^2 + 32}{4b - 1} = 4b + 1 + \frac{33}{4b - 1}$. This leads to $b = 1$ or $b = 3$, with the corresponding solutions being $(x, y) = (4, 2)$ or $(x, y) = (4, 6)$. Generally, $z^2a = \frac{z^2b^2 + z^3}{z^2b - 1} = b + \frac{b + z^3}{z^2b - 1}$. As a positive integer, $\frac{b + z^3}{z^2b - 1} \geq 1$, i.e., $b \leq \frac{z^2 - z + 1}{z - 1}$. When $z \geq 3$, $\frac{z^2 - z + 1}{z - 1} < z + 1$. Therefore, $b < z$. This implies $a \leq \frac{z^2 + z}{z^2 - 1} < 2$, so $a = 1$. Thus, $b$ is an integer solution to the equation $\omega^2 - z^2\omega + z + 1 = 0$. This indicates that the discriminant $z^4 - 4z - 4$ is a perfect square. However, it is strictly between $(z^2 - 1)^2$ and $(z^2)^2$, leading to a contradiction. Therefore, for $(x, y)$, only $(4, 2)$, $(4, 6)$, $(5, 2)$, and $(5, 3)$ are the solutions.