Actually, no. In any valuation domain, the prime ideals are linearly ordered by inclusion, so there exists at most one nonzero principal prime ideal.
In particular, let $K$ be any field, and let $$R=K[x,y/x,y/x^2,y/x^3,...],$$
i.e., elements of $R$ are "polynomials" in $x$ and $y$ over $K$, except you can divide $y$ by $x$ as many times as you like.
Consider the subset $S$ of $R$ containing all elements with nonzero constant term. It is clear that $S$ is a multiplicative subset of $R$, and let $T=R_S$ be the localization of $R$ at $S$--i.e., $$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f(x,y)\in R, g(x,y)\in S\right\}.$$
It isn't difficult to show that any nonzero element of $T$ is of the form $ux^n y^m$ where $u\in U(T)$, $m\geq 0$, and if $m=0$ then $n\geq 0$. (Basically, take an arbitrary nonzero element of $T$ and factor out all of the $x$'s and $y$'s that you can.)
So, we have the following chain of prime ideals in $T$ (and, in fact, these are all the prime ideals of $T$): $$0\subsetneq (y,y/x,y/x^2,y/x^3,\cdots)\subsetneq xT$$
You can generalize this to a chain of length $n$ by taking a valuation domain with value group isomorphic to $\mathbb{Z}^n$ under the lexicographic ordering.
Lemma: If $R$ is a Noetherian ring, and $P<R$ is a prime of height $n$, then there exist $x_1,\dots,x_n\in P$ such that $P$ is a minimal prime over $\langle x_1,\dots,x_n\rangle$.
Proof: By induction on $n$. The case $n=0$ is clear, since then $P$ is minimal over the zero ideal, which is generated by $\varnothing$. For the inductive step, suppose $n>0$. Now, since $R$ is Noetherian, it has finitely many minimal prime ideals; let $Q_1,\dots,Q_k$ be a complete list. Since $\operatorname{ht}Q_i=0<n=\operatorname{ht}P$ for each $i$, we have $P\nsubseteq Q_i$ for each $i$, and hence by prime avoidance we may find $x_n\in P\setminus\bigcup_{i=1}^nQ_i$. Then the prime ideal $P\big/\langle x_n\rangle$ of the Noetherian ring $R\big/\langle x_n\rangle$ has height strictly smaller than $n$, so by the inductive hypothesis there exist $x_1,\dots,x_{n-1}\in R$ such that $P\big/\langle x_n\rangle$ is a minimal prime over $\langle \bar{x}_1,\dots,\bar{x}_{n-1}\rangle$. Now $P$ is a minimal prime over $\langle x_1,\dots,x_n\rangle$, so we are done. $\blacksquare$
Now, for convenience denote $A=R[[X]]$. To show $\dim A\leqslant 1+\dim R$, it suffices to show $\operatorname{ht}M\leqslant 1+\dim R$ for any maximal ideal $M<A$, so let $M$ be a maximal ideal of $A$. As you have noted, we then have $M=\mathfrak{m}A+\langle X\rangle$, where $\mathfrak{m}=M\cap R$ is a maximal ideal of $R$. In particular, $k:=\operatorname{ht}\mathfrak{m}\leqslant\dim R$. By the lemma, there exist $r_1,\dots,r_k\in\mathfrak{m}$ such that $\mathfrak{m}$ is a minimal prime over $\langle r_1,\dots,r_k\rangle$. But now $M$ is a minimal prime over $\langle r_1,\dots,r_k,X\rangle$, so by Krull's height theorem $\operatorname{ht}M\leqslant k+1\leqslant 1+\dim R$, so we are done.
Best Answer
Hints:
$"\Rightarrow"$: Suppose that $R$ is a UFD and let $P \subset R$ be a prime ideal of height $1$. Then $P \neq 0$ such that we can take a non-zero $x \in P$ and consider its factorization into irreducibles. Can you now maybe deduce that one of these irreducibles generates $P$?
$"\Leftarrow"$: Suppose that every prime ideal of height $1$ is principal. Since $R$ is noetherian, every non-zero non-unit element $x$ has a factorization into irreducibles. It actually suffices to prove that an irreducible element $x$ is prime (why?). Now consider a minimal prime over $(x)$. This has height $1$ then and thus is generated by some single element $y$. Can you show that $(x) = (y)$ (which implies that $x$ is prime then)?