Answer
$
\mathbf{Z}^{+} \times \mathbf{Z}^{+}
$
is countable.
Work Step by Step
Let us define the function $f$ as:
$f : \mathbf{Z}^{+} \times \mathbf{Z}^{+} \rightarrow \mathbf{Z}^{+}, f(m, n)=2^{m} \cdot 3^{n}$
Check that $f$ is one-to-one: if $f(a,b)$=$f(m,n)$, then
$2^{a} \cdot 3^{b}=2^{m} \cdot 3^{n}$
Since $2$ and $3$ are premise:
$\begin{aligned} 2^{a} &=2^{m} \\ 3^{b} &=3^{n} \end{aligned}$
Since the bases are the same, the powers also have to be the same (else the positive integers cannot be the same).
$a=m$
$b=n$
By the definition of one-to-one, we have shown that$f$ is a one-to-one function.
Also as we know that there is a one-to-one function from $A$ to $B$ if and only if $$
|A| \leq|B|
$$
$$
\left|\mathbf{Z}^{+} \times \mathbf{Z}^{+}\right| \leq\left|\mathbf{Z}^{+}\right|
$$
A sert $A$ is countable if and only if $
|A| \leq\left|\mathbf{Z}^{+}\right|
$. Thus we have shown that $
\mathbf{Z}^{+} \times \mathbf{Z}^{+}
$
is countable.
