Its not the case that every UFD is noetherian; the standard counterexample is $R[x_0,x_1,x_2,\ldots]$, which has the following ascending sequence of ideals:
$$\langle \rangle,\langle x_0\rangle,\langle x_0,x_1\rangle,\langle x_0,x_1,x_2\rangle,\ldots$$
But notice that most of these ideals aren't principal. So:
Question. Is it the case that every UFD is noetherian on principal ideals?
I believe the answer is yes. Let $R$ be a UFD. Note that containment of two principal ideals $(a) \subset (b)$ corresponds with the division property $b \mid a$ for $a, b \in R$. Since $R$ is a UFD, every nonzero element has (up to multiplication by units) a unique finite factorization into irreducible elements, so let $a = x_{1}^{l_{1}}x_{2}^{l_{2}}\cdots x_{n}^{l_{n}}$ for irreducibles $x_{1}, \ldots, x_{n} \in R$ and some integers $l_{1}, \ldots, l_{n} \in \mathbb{N}$. Then $(a) \subset (b)$ implies that $b$ is a product of some subset of $\{x_{1}, \ldots, x_{n}\}$ raised to appropriate powers. It is then clear that every chain of principal ideals must eventually stabilize, since the length of any such chain beginning with $(a)$ is bounded by $\prod_{i=1}^{n} l_{i}$.