111 Determine all positive integers $x, y$ that satisfy $x+y^{2}+z^{3}=x y z$, where $z$ is the greatest common divisor of $x$ and $y$.
11. From the conditions, we can set \(x = za, y = zb\), where \(a, b\) are positive integers, and \((a, b) = 1\). Substituting into the equation, we get
$$a + z b^{2} + z^{2} = z^{2} a b$$

Thus, \(z \mid a\). Let \(a = zm\), then the equation becomes
$$m + b^{2} + z = z^{2} m b$$

Therefore, \(b \mid m + z\). Let \(m + z = bk\), then

Hence
$$\begin{aligned}
b^{2} + b k & = z^{2} m b \\
b + k & = z^{2} m
\end{aligned}$$

Notice that \(b k - (b + k) + 1 = (b - 1)(k - 1) \geqslant 0\),
Thus
$$b + k \leqslant b k + 1$$

Therefore
$$z^{2} m = b + k \leqslant b k + 1 = z + m + 1$$

Thus
$$\begin{array}{c}
m(z^{2} - 1) \leqslant z + 1 \\
m(z - 1) \leqslant 1 \\
z \leqslant 2
\end{array}$$

If \(z = 1\), then from (1) we have

Thus
$$\begin{array}{c}
m + b^{2} + 1 = m b \\
b^{2} - m b + (m + 1) = 0
\end{array}$$

This requires \(\Delta = m^{2} - 4(m + 1)\) to be a perfect square. Let \(m^{2} - 4(m + 1) = n^{2}\), where \(n\) is a non-negative integer, then
$$(m - n - 2)(m + n - 2) = 8$$

Solving this, we get
$$(m, n) = (5, 1)$$

Thus
$$b = 2, 3,$$

Therefore
$$(x, y, z) = (5, 2, 1), (5, 3, 1)$$

Thus
$$\begin{array}{c}
m(z - 1) \leqslant 1 \\
m = 1
\end{array}$$

Therefore, from (1) we get

Thus
$$\begin{array}{c}
b^{2} - 4 b + 3 = 0 \\
b = 1, 3
\end{array}$$

Therefore
$$(x, y, z) = (4, 2, 2), (4, 6, 2)$$

In summary, the solutions \((x, y)\) that satisfy the conditions are \((5, 2), (5, 3), (4, 2), (4, 6)\).