Lemma 10.119.8 in this page on Stacks project states that a polynomial ring over a UFD is a UFD.
It uses Nagata's criterion for factoriality (10.119.7): If $A$ is a domain and $S\subset A$ a multiplicative subset generated by prime elements, and $x\in A$ is irreducible, then it's image in $S^{-1}A$ is irreducible or a unit, and $x$ is prime if and only if it's image in $S^{-1}A$ is prime or a unit.
From this, it follows that $A$ is a UFD if and only if every element of $A$ factors into irreducibles and $S^{-1}A$ is a UFD.
Now, to prove Lemma 10.119.8, $S$ is taken to be generated by all prime elements of $A$ and so $S^{-1}A$ is the field of fractions of $A$, and therefore $S^{-1}(A[X])=(S^{-1}A)[X]$ is a UFD. From this and Nagata's criterion it follows that every irreducible element of $A[X]$ is prime.
However, it still remains to show that every element of $A[X]$ factors into irreducibles, and I don't see how the proof in the link I gave addresses this point.
One approach would be to take $p(X)\in A[X]$, factor it in $(S^{-1}A)[X]$, and lift the factorization to $s\cdot p(X)=f_1(X)\cdots f_r(X)$ where $s\in S$ and the $f_i(X)$ are polynomials in $A[X]$ which are irreducible in $(S^{-1}A)[X]$. However, I don't know how to continue without getting my hands dirty with coefficients (I think that one of the points of this proof is to avoid using results like Gauss lemma, etc.).
In fact, the way they state the Nagata's criterion is not the best possible, and you have experienced this on your own. (Of course, they wanted to give the most general frame for Nagata's criterion, but ...)
Replace the hard to check condition "every element factors into irreducibles" by the ascending chain condition on principal ideals (ACCP) (which implies the factorization into irreducibles) and show that $A$ has ACCP implies $A[X]$ has ACCP (for $A$ an integral domain).