The only integers that I've found to work are (12,6,4) and (10,10,2).

So if you assign these ordered triples to x,y, and z and cycle the values you should get 12 different integer combinations of x,y, and z that will solve the equations.

So as you can see there are \(\displaystyle 9 \) solutions.
However if you disregard order there are \(\displaystyle 2 \) solutions namely \(\displaystyle (x,y,z)=(12,6,4) \text{ and } (10,10,2) \), as oldguynewstudent has already pointed out.

Edit: I just realized you wanted your equation solved over \(\displaystyle \mathbb{Z} \), not \(\displaystyle \mathbb{N} \). This throws my idea right out the window...