We are tasked with finding all positive integer solutions $(x, y, z)$ to the system of equations:

1. $z + y^2 + x^3 = xyz$
2. $x = \gcd(y, z)$

First, observe that since $x = \gcd(y, z)$, $x$ divides both $y$ and $z$. Let us express $y$ and $z$ in terms of $x$:
$$y = xa \quad \text{and} \quad z = xb$$
where $\gcd(a, b) = 1$. Substituting these into the first equation gives:
$$xb + (xa)^2 + x^3 = (x)(xa)(xb)$$
Simplifying, this becomes:
$$xb + x^2 a^2 + x^3 = x^3 ab$$
Divide through by $x$ (assuming $x \neq 0$, which is valid for positive integers):
$$b + x a^2 + x^2 = x^2 ab$$
Rearranging this, we get:
$$b + x a^2 + x^2 = x^2 ab \quad \Rightarrow \quad b(1 - x^2 a) = x(a^2 + x)$$

To solve, consider small values of $x$ and find compatible $(y, z)$:

**Case $x = 1$:**

- For $x = 1$, the equations simplify to:
$$z + y^2 + 1 = yz$$
Rearrange to:
$$z - yz = -y^2 - 1 \quad \Rightarrow \quad z(1 - y) = -y^2 - 1$$
Try small values for $y$:
- $y = 2$:
$$z(1 - 2) = -4 - 1 \quad \Rightarrow \quad -z = -5 \quad \Rightarrow \quad z = 5$$
Valid solution: $(1, 2, 5)$.
- $y = 3$:
$$z(1 - 3) = -9 - 1 \quad \Rightarrow \quad -2z = -10 \quad \Rightarrow \quad z = 5$$
Valid solution: $(1, 3, 5)$.

**Case $x = 2$:**

- For $x = 2$:
$$z + y^2 + 8 = 2yz$$
Rearrange to:
z(2y - 1) = y^2 + 8 ] Trying small values of \( y \) gives: - \( y = 2 \): \[ z(2 \cdot 2 - 1) = 2^2 + 8 \quad \Rightarrow \quad 3z = 12 \quad \Rightarrow \quad z = 4
Valid solution: $(2, 2, 4)$.
- $y = 6$:
$$z(2 \cdot 6 - 1) = 6^2 + 8 \quad \Rightarrow \quad 11z = 44 \quad \Rightarrow \quad z = 4$$
Valid solution: $(2, 6, 4)$.

Hence, the solutions are:
$$\boxed{(1, 2, 5), (1, 3, 5), (2, 2, 4), (2, 6, 4)}$$

These exhaustively account for all solutions given the constraints.