Proving that a set is countable by finding a bijection 
$Z$ is the set of non-negative integers including $0$. Show that $Z \times Z \times Z$ is countable by constructing the actual bijection $f: Z\times Z\times Z \to \mathbb{N}$ ($\mathbb{N}$ is the set of all natural numbers). There is no need to prove that it is a bijection.

After searching for clues on how to solve this, I found $(x+y-1)(x+y-z)/z+y$ but that is only two dimensional and does not include $0$.  Any help on how to solve this?

If you don't need an actual "formula", then you can write
$$
\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z} = \bigcup_{n=0}^\infty \{ (x,y,z)\in \mathbb{Z}^3 : |x|+|y| +|z| = n \}
$$
and then rely on the fact that each term in this union is a finite set.
