To do this proof I'd like to get the conclusion from the title from an easier statement: $(a)(b)=(ab)$ for $a,b\in R$. If I'm right, in a commutative ring the principal ideal $(a)$ is generated by $\{ca:c\in R\}$ and $\{db:d\in R\}$.
Proving $(a)(b)\subseteq(ab)$: If $x\in (a)(b)$ is $x=ca\cdot db=cdab$ for some $c,d\in R$, $cd\in R$ then $cd\cdot ab\in(ab)\implies x\in (ab)\implies (a)(b)=(ab)$.
Proving $(ab)\subseteq(a)(b)$. If $x\in(ab)$ is $x=c\cdot ab=[c\cdot a][e\cdot b]$ where the first is in $(a)$ and the second in $(b)$, then $x\in (a)(b)$.
Is this right?. Then to prove $(m\mathbb{Z})(n\mathbb{Z})=mn\mathbb{Z}$ take $a=m,b=n$ just like in the former case and the elements of $(m\mathbb{Z})(n\mathbb{Z})$ can be written as elements of $mn\mathbb{Z}$ and conversely.
Should I add something to be more formal?.
It's not completely right, but the idea is good.
If $x\in(a)(b)$, then $$ x=\sum_{i=1}^n y_ia\cdot z_ib $$ for some $y_1,\dots,y_n,z_1,\dots,z_n\in R$. Then $$ x=\biggl(\sum_{i=1}^n y_iz_i\biggr)ab\in (ab) $$ Or, perhaps more simply, $(a)(b)$ is (additively) generated by the elements of the form $yz$, with $y\in(a)$ and $z\in(b)$. But for such an element we have $y=ca$ and $z=db$, so $yz=cd\cdot ab\in(ab)$. Hence every generating element of $(a)(b)$ lives in $(ab)$ and therefore $(a)(b)\subseteq(ab)$.
The inclusion $(a)(b)\subseteq(ab)$ requires the assumption that the ring has unit: if $x\in(ab)$, then $x=c\cdot ab=ca\cdot 1b\in(a)(b)$.