Proposition: Let $a,b \in R$. If $p(X),q(X)\in R[X]$ are primitive and $R$ is an integral domain, then $ap(X)=bq(X)$ imply $a$ and $b$ are associates; so are $p(X)$ and $q(X)$.
Proof:
Let \begin{align} p(X) &= p_mX^m + p_{m-1}X^{m-1} + \cdots + p_1X + p_0\\ q(X) &= q_nX^n + q_{n-1}X^{n-1} + \cdots + q_1X + q_o \end{align} By definition of a primitive, \begin{align} \forall r, \overline{r} \in R ((r \mid p_0, p_1, \cdots, p_m \quad \land \quad \overline{r} \mid q_0, q_1,\cdots, q_n)\Leftrightarrow (\text{$r$ and $\overline{r}$ are units of $R$.})). \end{align} Suppose $ap(X)=bq(X)$, then $m=n$ and $\forall i=0,1, \cdots, n(ap_i=bq_i)$. It follows, $a \mid bq_0, bq_1, \cdots, bq_n$ and $b\mid ap_0,ap_1, \cdots, ap_n$. Note that $a$ and $b$ are either prime or not prime. If either one of them are prime, then, by Lemma 3, $b \nmid p_0, p_1, \cdots, p_n$. It follows, $a=bu$ and since primes are irreducible, $u \in R$ is a unit. Hence, $a$ and $b$ are associates. Also, by substitution of $a$ with $bu$ and cancellation of $b$, $p(X)$ and $q(X)$ are associates.
I don't know how to proceed when neither $a$ nor $b$ are prime. :(