Solve in $ \mathbb{Z} $ the following system of equations:
$$ \left\{\begin{matrix} 5^x-\log_2 (y+3) = 3^y\\ 5^y -\log_2 (x+3)=3^x\end{matrix}\right. . $$
1. Consider the given system of equations:
   \[
   \begin{cases}
   5^x - \log_2 (y+3) = 3^y \\
   5^y - \log_2 (x+3) = 3^x
   \end{cases}
   \]

2. Suppose \( x > y \). Then \( 3^x > 3^y \). This implies:
   \[
   5^y + \log_2 (y+3) > 5^x + \log_2 (x+3)
   \]
   This is a contradiction because if \( x > y \), then \( 5^x > 5^y \) and \( \log_2 (x+3) > \log_2 (y+3) \). Therefore, we must have \( x = y \).

3. Substituting \( x = y \) into the system, we get:
   \[
   5^x - \log_2 (x+3) = 3^x
   \]
   Simplifying, we obtain:
   \[
   5^x = 3^x + \log_2 (x+3)
   \]

4. To solve this equation, consider the function:
   \[
   f(x) = 5^x - 3^x - \log_2 (x+3)
   \]
   We need to find \( x \) such that \( f(x) = 0 \).

5. Evaluate \( f(x) \) at \( x = 1 \):
   \[
   f(1) = 5^1 - 3^1 - \log_2 (1+3) = 5 - 3 - \log_2 4 = 5 - 3 - 2 = 0
   \]
   Thus, \( x = 1 \) is a solution.

6. To determine if there are any other solutions, consider the behavior of \( f(x) \). The function \( 5^x \) grows exponentially, \( 3^x \) also grows exponentially but at a slower rate, and \( \log_2 (x+3) \) grows logarithmically. Therefore, \( 5^x \) will dominate the other terms as \( x \) increases.

7. The function \( f(x) \) is monotonically increasing for \( x > 1 \) because the derivative \( f'(x) = 5^x \ln 5 - 3^x \ln 3 - \frac{1}{(x+3) \ln 2} \) is positive for \( x > 1 \). This implies that \( f(x) \) can have at most one root.

8. Since \( x = 1 \) is a root and \( f(x) \) is monotonically increasing for \( x > 1 \), there are no other solutions.

The final answer is \( \boxed{ x = 1 } \) and \( y = 1 \).