Let $f, g \in \mathbb{Q}[x]$ with $fg\in\mathbb{Z}[x]$. A proof I found is the following:
Let $f(x) = a_nx^n+ \dots + a_1 x + a_0$, and $g(x) = b_mx^m+\dots+b_1x+b_0$ where $a_i, b_i \in \mathbb{Q}$. By Gauss’ Lemma, if the product $f(x)g(x)$ factors in $\mathbb Q[x]$, then there exist $r, s \in Q$ so that $rf(x), sg(x) \in \mathbb{Z}[x]$ and $rs = 1$. Hence, for any $i, j, a_ib_j = (rs)a_ib_j = (ra_i)(sb_j ) \in \mathbb{Z}$.
However, I am confused why "By Gauss’ Lemma, if the product $f(x)g(x)$ factors in $\mathbb Q[x]$, then there exist $r, s \in \mathbb Q$ so that $rf(x), sg(x) \in \mathbb{Z}[x]$ and $rs = 1$." Since we have that $fg$ is reducible in $\mathbb{Q}[x]$, doesn't that mean we only know that $fg$ is reducible in $\mathbb{Z}[x]$ by Gauss' Lemma? Why do we know that there exist $r, s \in Q$ so that $rf(x), sg(x) \in \mathbb{Z}[x]$ and $rs = 1$? This seems like a stronger result.
It depends on which version of the Gauß Lemma you take. I do not think that the one you stated is strong enough to prove what you want to show. Your version is a Corollary of the following:
Lemma of Gauß: Let $f, g \in \mathbb{Q}[x]$. Then $cont(fg) = cont(f) \cdot cont(g)$.
For the definition of content, see https://en.m.wikipedia.org/wiki/Primitive_part_and_content .
Now in your situation, if we write $a=cont(f)$ and $b=cont(g)$, we have by the Lemma above that $ab=cont(fg) \in \mathbb{Z}$. By the definition of content, we have $f/a, g/b \in \mathbb{Z}[x]$. Now set $r=1/a$ and $s = a$. Then we have $rs=1$, $rf \in \mathbb{Z}[x]$ and $sg = ag = ab \cdot g/b \in \mathbb{Z}[x]$.