Let $f$ and $g$ be monic polynomials in $\mathbb Z[x]$.
Then, \begin{align} \operatorname{disc} (f\cdot g) = \operatorname{disc}(f)\cdot \operatorname{disc}(g) \cdot \prod_i\prod_j(a_i-b_j)^2 , \end{align} where the $a_i$ are the roots of $f$ (with multiplicities) and the $b_j$ are the roots of $g$ (with multiplicities).
It seems that the term $\prod_i\prod_j(a_i-b_j)^2$ is always a square in $\mathbb Z$, but I can't rewrite it to see this.
The task is equivalent to showing that $\prod_{i,j} (a_i-b_j)$ is an integer. This follows from the fact that the expression is symmetric in the $a_i$ and symmetric in the $b_j$.
Explicitly, consider the corresponding polynomial $F(x) = \prod_{i,j} (x_i-y_j)$ in $\mathbb Z[x_1, \ldots, x_m, y_1, \ldots, y_n]$, where $m = \deg f, n = \deg g$. If you treat $F$ as an element of $R[y_1, \ldots, y_n]$, where $R = \Bbb Z[x_1, \ldots, x_m]$, then the result is evidently symmetric, in that any permutation of the $y_j$ merely permutes terms of $F$.
Using the fundamental theorem of symmetric polynomials, we can rewrite it as a polynomial in the elementary symmetric polynomials in the $y_j$ over $R$. Note that up to sign these are the coefficients of $g$, and are thus integers. This means that if we set $y_j = b_j$, we get an element of $\Bbb Z[x_1, \ldots, x_m]$. This is similarly symmetric in the $x_i$, so we can plug in $x_i = a_i$ and get an integer at the end by the same argument.