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Consider the ideal $I=(2,x)$ in $\mathbb{Z}[x]$. It is proper: evaluation at zero followed by reduction modulo $2$ maps the ring onto ${\mathbb{F}}_2$ and has kernel $I$. Suppose $I=(f)$. Then $f$ divides the constant polynomial $2$, so $f$ is a constant integer dividing $2$. Up to sign, the only possibilities are $1$ and $2$. The first would make the ideal the whole ring; the second cannot divide $x$ in $\mathbb{Z}[x]$. Both are impossible. This proves the integer polynomial ring is not a principal ideal domain result: $\mathbb{Z}[x]$ is not a principal ideal domain.

For the same pair, $\boxed{S=\mathbb{Z}[X]\subset \mathbb{Q}[X]}$. Since $\mathbb{Z}$ is a unique factorization domain, Gauss lemma for polynomials proves that $S$ is a unique factorization domain.

To prove that $S$ is not a principal ideal domain, consider the ideal $I=(2,X)$. A generator $g$ would divide 2 and $X$. Divisibility of 2 forces $g$ to be a nonzero constant integer; divisibility of $X$ then forces that integer to divide its coefficient 1, so $g=\pm 1$. This would make $I=S$. But evaluation at $X=0$ followed by reduction modulo 2 maps every element of $I$ to zero and maps 1 to one. Hence $1\notin I$, a contradiction. This is the integer polynomial ring is not a principal ideal domain example, with both clauses satisfied by a single inclusion.