Proof $\mathbb{Z} \times \mathbb{Z}$ is cyclic.
My intuition is that there are four generators in $\mathbb{Z} \times \mathbb{Z}: 1 \times 1, 1 \times -1, -1 \times 1, -1 \times -1$. And the group they generated are $(z,z)$ or $(z,-z)$, rather than two independent $(z_1, z_2)$, where $z_1, z_2 \in \mathbb{Z}$.
But is this the right way to think about it, and right way to write it? Thank you. :-)
You’re attempting the impossible. The subgroup generated by $\langle m,n\rangle\in\Bbb Z\times\Bbb Z$ is $$\{\langle mk,nk\rangle:k\in\Bbb Z\}\;;$$ can you find any pair $\langle m,n\rangle\in\Bbb Z\times\Bbb Z$ for which this set is all of $\Bbb Z\times\Bbb Z$?
Any quotient of a cyclic group is cyclic but $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is a non-cyclic quotient of $\mathbb{Z} \times \mathbb{Z}$.
$\Bbb{Z} \times \Bbb{Z}$ is not cyclic. You've correctly realized that if it were cyclic, then its generator would have to have a very specific form; but it's routine to see that none of the elements you've listed is a generator.
For a more formal way of writing, suppose that some $(a, b)$ were a generator. It's clear that $|a| \le 1$ and $|b| \le 1$ (else, we'd miss elements of the form $(0, 1)$ and $(1, 0)$). It's also clear that $a \ne 0$ and $b \ne 0$.
Now note that $(a, -b) \notin \langle (a, b) \rangle$ and we're done.