Solve in $ \mathbb{Z} $ the system:
$$ \left\{  \begin{matrix} 2^x+\log_3 x=y^2 \\ 2^y+\log_3 y=x^2 \end{matrix} \right. . $$
1. **Define the function \( f(t) \):**
   Consider the function \( f(t) = 2^t + \log_3 t \). We need to analyze the behavior of this function to solve the given system of equations.

2. **Analyze the monotonicity of \( f(t) \):**
   To determine if \( f(t) \) is increasing for \( t \geq 1 \), we compute its derivative:
   \[
   f'(t) = \frac{d}{dt} \left( 2^t + \log_3 t \right) = 2^t \ln 2 + \frac{1}{t \ln 3}
   \]
   For \( t \geq 1 \), both \( 2^t \ln 2 \) and \( \frac{1}{t \ln 3} \) are positive. Therefore, \( f'(t) > 0 \) for \( t \geq 1 \), which implies that \( f(t) \) is strictly increasing for \( t \geq 1 \).

3. **Equate the two functions:**
   Given the system of equations:
   \[
   \begin{cases}
   2^x + \log_3 x = y^2 \\
   2^y + \log_3 y = x^2
   \end{cases}
   \]
   Since \( f(t) \) is strictly increasing for \( t \geq 1 \), if \( x \geq 1 \) and \( y \geq 1 \), then \( f(x) = y^2 \) and \( f(y) = x^2 \) imply that \( x = y \).

4. **Substitute \( x = y \) into the equation:**
   Substituting \( x = y \) into either equation, we get:
   \[
   2^x + \log_3 x = x^2
   \]
   We need to solve this equation for \( x \) in \( \mathbb{Z} \).

5. **Solve the equation \( 2^x + \log_3 x = x^2 \):**
   Let's test some integer values of \( x \):
   - For \( x = 1 \):
     \[
     2^1 + \log_3 1 = 1 + 0 = 1 \neq 1^2
     \]
   - For \( x = 2 \):
     \[
     2^2 + \log_3 2 = 4 + \log_3 2 \approx 4 + 0.631 \approx 4.631 \neq 2^2
     \]
   - For \( x = 3 \):
     \[
     2^3 + \log_3 3 = 8 + 1 = 9 = 3^2
     \]
     This satisfies the equation.

6. **Verify the solution:**
   If \( x = 3 \), then \( y = 3 \). Substituting back into the original system:
   \[
   \begin{cases}
   2^3 + \log_3 3 = 9 \\
   2^3 + \log_3 3 = 9
   \end{cases}
   \]
   Both equations are satisfied.

Conclusion:
The only solution in \( \mathbb{Z} \) is \( x = 3 \) and \( y = 3 \).

The final answer is \( \boxed{ (x, y) = (3, 3) } \)