let $R$ be a commutative ring with unit, and fix elements $A(X)$ and $B(X)$, not both zero, in $R[X]$.
suppose that there exist polynomials $s(X)$ and $t(X)$ in $R[X]$, respectively of degrees less than $\deg(B)$ and $\deg(A)$, and not both zero, such that $s(X) \cdot A(X) + t(X) \cdot B(X) = 0$. (the existence of such $s$ and $t$ is implied by the vanishing of the resultant $\text{res}(A, B)$.)
Question. Can we conclude that $A$ and $B$ have a nontrivial common divisor $g$ in $R[X]$? (i.e., of positive degree, and hopefully monic?)
when $R$ is a field, this is a standard aspect of the theory of resultants, and essentially follows from unique factorization: let's assume wlog that $t \neq 0$. the relation implies $A \mid t \cdot B$; if $A$ and $B$ were coprime, then $A \mid t$ would hold, contradicting $\deg(t) < \deg(A)$.
Even with $R$ a Dedekind Domain your property need not hold. For example, let $R=\mathbb{Z}[\sqrt{-5}]$:
Let $$A(X)=(1+\sqrt{-5})X+2,\qquad\qquad\qquad$$ $$B(X)=(1+\sqrt{-5})X^2+5X+1-\sqrt{-5}.\,$$
Then $$(2x+1-\sqrt{-5})A(x)=2B(X).$$
However $A(X)$ and $B(X)$ cannot have any common linear factor, as $A$ is not divisible by any non-unit in $R$, and $A$ does not divide $B$.
Conversely, starting from your assertion that the result holds for fields, it is easy to show that it holds for $R$ a UFD:
Passing to the field of fractions of $R$, we obtain a polynomial $f(X)$ (over $R$) of degree at least $1$, such that $$f(X)\,|\,\lambda A(X),\qquad\qquad f(X)\,|\,\mu B(X),$$
with $\lambda,\mu \in R$ both non-zero.
For any prime factor $p$ of $\lambda$, we have $$p\,|\,f(X)g(X)=\lambda A(X)$$ for some polynomial $g(X)$. Thus either $p$ divides $f(X)$, so
$$f(X)/p\,|\,\lambda/p A(X),\qquad\qquad f(X)/p\,|\,\mu B(X),$$
or $p$ divides $g(X)$, so
$$f(X)\,|\,\lambda/p A(X),\qquad\qquad f(X)\,|\,\mu B(X),$$
Thus we may remove all the prime factors from $\lambda$ and $\mu$ one by one, till we are left with $$f'(X)\,|\, A(X),\qquad\qquad f'(X)\,|\, B(X),$$ for some polynomial $f'(X)$ of degree at least $1$.