It is known that a valuation domain is a principal ideal ring if and only if its prime ideals are principal. Is it a principal ideal ring when its unique maximal ideal is principal?

1$\begingroup$ Yes, a proof can be found (for instance) in Serre's Local Fields. $\endgroup$ – Torsten Ekedahl Aug 25 '10 at 4:01

$\begingroup$ @Torsten Ekedahl: could you tell me where can I find the proof in Serre's book? Thanks. $\endgroup$ – TmobiusX Aug 25 '10 at 6:38

2$\begingroup$ As Pete's example shows I forgot that the ring needs to be Noetherian. In any case if you can live with that restriction the result in Serre I:2.2. It is also true if the valuation is of rank 1 (which was I think was what really confused me). $\endgroup$ – Torsten Ekedahl Aug 25 '10 at 19:47
I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x \in K^{\times}$, at least one of $x,x^{1}$ lies in $R$.
In this case, I believe the answer is no. Let $R$ be any valuation domain whose value group $K^{\times}/R^{\times}$ is isomorphic, as a totally ordered abelian group, to $\mathbb{Z} \times \mathbb{Z}$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal.
For some information on valuation rings, see e.g. Section 17 of