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?

Just for fun, here’s an explicit enumeration.
First do it with $\Bbb N\times\Bbb N\times\Bbb N$. If you think of this as a subset of $\Bbb R^3$, you can chop it up into points lying in the parallel planes $P_k$ defined by $x+y+z=k$ for $k\in\Bbb N$. $P_0$ contains only $\langle 0,0,0\rangle$, $P_1$ contains $\langle 0,0,1\rangle,\langle 0,1,0\rangle$, and $\langle 1,0,0\rangle$ and so on. There are $\binom{k+2}2$ solutions to $x+y+z=k$ in non-negative integers $x,y$, and $z$, so $P_k$ contains $\binom{k+2}2$ members of $\Bbb N^3$. We’ll enumerate $\Bbb N^3$ by enumerating the individual $P_k$ in order of increasing $k$. We’ll start by explicitly enumerating the $\binom{k+2}2$ points of $P_k$.
Suppose that $\langle x,y,z\rangle\in P_k$. Then $x+y=k-z$, and a little thought shows that there are only $k-z+1$ possibilities, namely $x=i$ and $y=k-x-i$ for $i=0,\dots,k-z$. We’ll start by listing $P_k$ in reverse order of $z$. For $z=k$ there is only one point, $\langle 0,0,k\rangle$; in this ordering it has $0$ predecessors in $P_k$. For $z=k-1$ there are two points, $\langle 0,1,k-1\rangle$ and $\langle 1,0,k-1\rangle$; we’ll list them in increasing order of $x$, so that $\langle 0,1,k-1\rangle$ has $1$ predecessor, and $\langle 1,0,k-1\rangle$ has $2$ predecessors. For $z=k-2$ there are $3$ points; again listing them in increasing order of $x$, we have $\langle 0,2,k-2\rangle$ with $3$ predecessors, $\langle 1,1,k-2\rangle$ with $4$ predecessors, and $\langle 2,0,k-2\rangle$ with $5$ predecessors.
In general, $\langle 0,i,k-i\rangle$ has $\sum_{j=1}^ij=\binom{i+1}2$ predecessors, so $\langle j,i-j,k-i\rangle$ has $\binom{i+1}2+j$ predecessors. Thus, if $\langle x,y,z\rangle\in P_k$, it has $\binom{k-z+1}2+x$ predecessors in this ordering of $P_k$.
Next, note that $$\sum_{0\le i<k}|P_i|=\sum_{0\le i<k}\binom{i+2}2=\binom{k+2}3\;.$$ Thus, in the ordering of $\Bbb N^3$ as a whole, $\langle 0,0,k\rangle$ has $\binom{k+2}3$ predecessors. Combining results, we see that $\langle x,y,z\rangle$ has $$\binom{x+y+z+2}3+\binom{x+y+1}2+x$$ predecessors in this ordering of $\Bbb N^3$. (That last term can be thought of as $\binom{x+0}1$, and the obvious pattern can be generalized to higher powers of $\Bbb N$.) Since an enumeration $X\to\Bbb N$ of a set $X$ is simply a function that assigns to each $x\in X$ the number of predecessors of $x$ in the induced ordering, we have our enumeration of $\Bbb N^3$:
$$\varphi:\Bbb N^3\to\Bbb N:\langle x,y,z\rangle\mapsto\binom{x+y+z+2}3+\binom{x+y+1}2+x\;.$$
All that remains is to convert this to an enumeration of $\Bbb Z^3$. This is easily done by starting with the enumeration $$\psi:\Bbb Z\to\Bbb N:n\mapsto(-1)^n\left\lceil\frac{n}2\right\rceil\;,$$ which orders $\Bbb Z$ as $\langle 0,-1,1,-2,2,-3,3,\dots\rangle$. Let $$\overline\psi:\Bbb Z^3\to\Bbb N^3:\langle x,y,z\rangle\mapsto\langle \psi(x),\psi(y),\psi(z)\rangle\;;$$ this is clearly a bijection, and the desired explicit enumeration $\sigma:\Bbb Z^3\to\Bbb N$ is now given by $\sigma=\varphi\circ\overline\psi$. That is, $\langle x,y,z\rangle\in\Bbb Z^3$ is sent to 
$$\begin{align*}
&\binom{(-1)^x\left\lceil\frac{x}2\right\rceil+(-1)^y\left\lceil\frac{y}2\right\rceil+(-1)^z\left\lceil\frac{z}2\right\rceil+2}3+\\
&\qquad\qquad+\binom{(-1)^x\left\lceil\frac{x}2\right\rceil+(-1)^y\left\lceil\frac{y}2\right\rceil+1}2+(-1)^x\left\lceil\frac{x}2\right\rceil\;.
\end{align*}$$
