A Noetherian ring with Krull dimension one which is not a Dedekind domain

The ring $\mathbb{Z}[2i]$ is an example. It satisfies all the properties of a Dedekind domain except that it is not integrally closed. (To see that it satisfies these properties, note that it is integral over $\mathbb{Z}$, so the Krull dimension is one, as integral extensions preserve dimension. It is also clearly noetherian.)

The ring $R_1 = \mathbb{Z} \times \mathbb{Z}$ gives a counterexample to your claim (it is not a domain). However $\operatorname{Spec}(R_1)$ is a Dedekind scheme, so this is a somewhat cheap counterexample. The counterexample $R_2 = \mathbb{Z}[t]/(t^2)$ is more serious.

If by "other near misses" you mean other rings that satisfy two of the three conditions for being a Dedekind domain, another such ring is k[x, y], which is Noetherian (by the Hilbert basis theorem), integrally closed (it is a UFD, and UFDs are integrally closed) but not every prime ideal is maximal (for example (x)).