The question is whether such a thing exists. Namely, a discrete valuation ring (DVR), in whatever way you define it, is quite obviously a domain, Noetherian, and has a unique prime element up to associates (i.e up to multiplying with units). The question is if there are rings which satisfy all these properties at once, but are not DVRs.
Question: Is there a Noetherian domain $R$ such that, besides $(0)$, there is exactly one principal prime ideal $(p)$ in $R$, but $R$ is not a DVR?
To clarify, I am not supposing that $R$ has a unique prime ideal, which happens to be principal. (In this case, it would be clear that $R$ is a DVR.) I suppose that $R$ has only one non-trivial ideal which is both prime and principal, while it might (should!) have other prime ideals, just that those are not principal.
This arose out of this question which asked generally for examples of commutative rings with only one prime element up to associates. All answers besides DVRs given there are either not Noetherian, or not domains, or neither.
There is of course a lot of theory about prime ideals in Noetherian domains, but I see very little which distinguishes between principal and non-principal such prime ideals, which is kind of what seems to be needed fur this little curious question.
Best Answer
For $char(k)\ne 2$ $$R=k[t^2,t^3]_{(t^2,t^3)}\cap k[t^2,t^3]_{(t^2-1,t^3-1)}$$ is not a DVR,
it has 3 prime ideals $$\{0\} \qquad (t^2,t^3) \qquad (t^2-1,t^3-1)$$
$(t^2,t^3)$ is not principal
$t^3+1\in R^\times$ so $t^3-1\in (t^6-1)=(t^2-1)(1+t^2+t^4)\subset (t^2-1)$ and thus $(t^2-1,t^3-1)=(t^2-1)$ is principal
It is Noetherian because it is a localization of $k[t^2,t^3]$ which is Noetherian.