Determine all trios of integers $(x, y, z)$ which are solution of system of equations
$\begin{cases} x - yz = 1 \\ xz + y = 2 \end{cases}$


We are given the system of equations:
\[
\begin{cases}
x - yz = 1 \\
xz + y = 2
\end{cases}
\]

1. **Express \( x \) in terms of \( y \) and \( z \) from the first equation:**
   \[
   x - yz = 1 \implies x = yz + 1
   \]

2. **Substitute \( x = yz + 1 \) into the second equation:**
   \[
   (yz + 1)z + y = 2
   \]
   Simplify the equation:
   \[
   yz^2 + z + y = 2
   \]

3. **Rearrange the equation to isolate terms involving \( y \):**
   \[
   yz^2 + y + z = 2 \implies y(z^2 + 1) + z = 2
   \]

4. **Solve for \( y \):**
   \[
   y(z^2 + 1) = 2 - z \implies y = \frac{2 - z}{z^2 + 1}
   \]

5. **Analyze the possible integer values for \( z \) such that \( y \) is an integer:**
   - For \( y \) to be an integer, \( \frac{2 - z}{z^2 + 1} \) must be an integer.

6. **Check possible values of \( z \):**
   - **If \( z = 2 \):**
     \[
     y = \frac{2 - 2}{2^2 + 1} = \frac{0}{5} = 0
     \]
     Substitute \( z = 2 \) and \( y = 0 \) back into \( x = yz + 1 \):
     \[
     x = 0 \cdot 2 + 1 = 1
     \]
     Thus, one solution is \( (x, y, z) = (1, 0, 2) \).

   - **If \( z = -1 \):**
     \[
     y = \frac{2 - (-1)}{(-1)^2 + 1} = \frac{3}{2}
     \]
     Since \( y \) is not an integer, \( z = -1 \) is not a valid solution.

   - **If \( z = 0 \):**
     \[
     y = \frac{2 - 0}{0^2 + 1} = \frac{2}{1} = 2
     \]
     Substitute \( z = 0 \) and \( y = 2 \) back into \( x = yz + 1 \):
     \[
     x = 2 \cdot 0 + 1 = 1
     \]
     Thus, another solution is \( (x, y, z) = (1, 2, 0) \).

   - **If \( z > 2 \):**
     \[
     y = \frac{2 - z}{z^2 + 1}
     \]
     For \( z > 2 \), \( 2 - z \) is negative and \( z^2 + 1 \) is positive, making \( y \) a non-integer.

   - **If \( z < -1 \):**
     \[
     y = \frac{2 - z}{z^2 + 1}
     \]
     For \( z < -1 \), \( 2 - z \) is positive and \( z^2 + 1 \) is positive, but the fraction does not simplify to an integer.

Therefore, the only integer solutions are:
\[
(x, y, z) \in \{(1, 0, 2), (1, 2, 0)\}
\]

The final answer is \(\boxed{(x, y, z) \in \{(1, 0, 2), (1, 2, 0)\}}\)