Solve in $ \mathbb{Z}^2 $ the following equation:
$$ (x+1)(x+2)(x+3) +x(x+2)(x+3)+x(x+1)(x+3)+x(x+1)(x+2)=y^{2^x} . $$


[i]Adrian Zanoschi[/i]
1. Start by expanding the left-hand side of the given equation:
   \[
   (x+1)(x+2)(x+3) + x(x+2)(x+3) + x(x+1)(x+3) + x(x+1)(x+2)
   \]

2. Expand each term individually:
   \[
   (x+1)(x+2)(x+3) = (x+1)(x^2 + 5x + 6) = x^3 + 5x^2 + 6x + x^2 + 5x + 6 = x^3 + 6x^2 + 11x + 6
   \]
   \[
   x(x+2)(x+3) = x(x^2 + 5x + 6) = x^3 + 5x^2 + 6x
   \]
   \[
   x(x+1)(x+3) = x(x^2 + 4x + 3) = x^3 + 4x^2 + 3x
   \]
   \[
   x(x+1)(x+2) = x(x^2 + 3x + 2) = x^3 + 3x^2 + 2x
   \]

3. Sum these expanded terms:
   \[
   (x^3 + 6x^2 + 11x + 6) + (x^3 + 5x^2 + 6x) + (x^3 + 4x^2 + 3x) + (x^3 + 3x^2 + 2x)
   \]
   \[
   = 4x^3 + (6x^2 + 5x^2 + 4x^2 + 3x^2) + (11x + 6x + 3x + 2x) + 6
   \]
   \[
   = 4x^3 + 18x^2 + 22x + 6
   \]

4. The equation now becomes:
   \[
   4x^3 + 18x^2 + 22x + 6 = y^{2^x}
   \]

5. We need to find integer solutions \((x, y)\) such that the above equation holds. Let's analyze the equation for small integer values of \(x\):

   - For \(x = 0\):
     \[
     4(0)^3 + 18(0)^2 + 22(0) + 6 = 6
     \]
     \[
     y^{2^0} = y^1 = y
     \]
     \[
     6 = y \implies y = 6
     \]
     Thus, \((0, 6)\) is a solution.

   - For \(x = 1\):
     \[
     4(1)^3 + 18(1)^2 + 22(1) + 6 = 4 + 18 + 22 + 6 = 50
     \]
     \[
     y^{2^1} = y^2
     \]
     \[
     50 = y^2 \implies y = \pm \sqrt{50} \implies y \notin \mathbb{Z}
     \]
     Thus, there are no integer solutions for \(x = 1\).

   - For \(x = 2\):
     \[
     4(2)^3 + 18(2)^2 + 22(2) + 6 = 4(8) + 18(4) + 22(2) + 6 = 32 + 72 + 44 + 6 = 154
     \]
     \[
     y^{2^2} = y^4
     \]
     \[
     154 = y^4 \implies y = \pm \sqrt[4]{154} \implies y \notin \mathbb{Z}
     \]
     Thus, there are no integer solutions for \(x = 2\).

   - For \(x = 3\):
     \[
     4(3)^3 + 18(3)^2 + 22(3) + 6 = 4(27) + 18(9) + 22(3) + 6 = 108 + 162 + 66 + 6 = 342
     \]
     \[
     y^{2^3} = y^8
     \]
     \[
     342 = y^8 \implies y = \pm \sqrt[8]{342} \implies y \notin \mathbb{Z}
     \]
     Thus, there are no integer solutions for \(x = 3\).

6. For \(x \geq 4\), the left-hand side grows much faster than the right-hand side, making it impossible for the equation to hold for integer \(y\).

Conclusion:
The only integer solution in \(\mathbb{Z}^2\) is \((0, 6)\).

The final answer is \( \boxed{ (0, 6) } \).