How many integer triples $(x,y,z)$ are there satisfying $x^3+y^3=x^2yz+xy^2z+2$ ?

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$
1. Start with the given equation:
   \[
   x^3 + y^3 = x^2yz + xy^2z + 2
   \]

2. Factor the left-hand side using the sum of cubes formula:
   \[
   x^3 + y^3 = (x+y)(x^2 - xy + y^2)
   \]

3. Rewrite the right-hand side:
   \[
   x^2yz + xy^2z + 2 = xyz(x + y) + 2
   \]

4. Equate the factored form of the left-hand side to the right-hand side:
   \[
   (x+y)(x^2 - xy + y^2) = xyz(x + y) + 2
   \]

5. Simplify by dividing both sides by \(x + y\) (assuming \(x + y \neq 0\)):
   \[
   x^2 - xy + y^2 = xyz + \frac{2}{x+y}
   \]

6. Multiply both sides by \(x + y\) again to clear the fraction:
   \[
   (x + y)(x^2 - xy + y^2) = 2
   \]

7. Expand and simplify:
   \[
   (x + y)(x^2 - xy + y^2 + xyz) = 2
   \]

8. Consider the possible values for \(x + y\):
   - If \(x + y = 2\):
     \[
     2(x^2 - xy + y^2 + xyz) = 2 \implies x^2 - xy + y^2 + xyz = 1
     \]
     Simplify:
     \[
     xy(z - 3) = -3
     \]
     Possible solutions:
     \[
     (1, 1, 0), (3, -1, 4), (-1, 3, 4)
     \]

   - If \(x + y = -2\):
     \[
     -2(x^2 - xy + y^2 + xyz) = 2 \implies x^2 - xy + y^2 + xyz = -1
     \]
     Simplify:
     \[
     xy(z - 3) = -5
     \]
     Possible solution:
     \[
     (-1, -1, 2)
     \]

   - If \(x + y = 1\):
     \[
     1(x^2 - xy + y^2 + xyz) = 2 \implies x^2 - xy + y^2 + xyz = 2
     \]
     Simplify:
     \[
     xy(z - 3) = 1
     \]
     No integer solutions.

   - If \(x + y = -1\):
     \[
     -1(x^2 - xy + y^2 + xyz) = 2 \implies x^2 - xy + y^2 + xyz = -2
     \]
     Simplify:
     \[
     xy(z - 3) = 3
     \]
     No integer solutions.

9. Summarize the integer solutions:
   \[
   (1, 1, 0), (3, -1, 4), (-1, 3, 4), (-1, -1, 2)
   \]

The final answer is \(\boxed{4}\)