Yes. In fact, there's an example of such a ring that is also a Bezout domain (so all the finitely generated primes are in fact principal). The construction is as follows. Fix an integer prime $p$, let $\alpha$ be an element of $\mathbb Z_p$ that is transcendental over $\mathbb Q$, and set $A = \{ \alpha + p, \alpha + p^2, \ldots \}$. The desired ring is then $R = \{ f \in \mathbb Q(X) \colon f(A) \subset \mathbb Z_p \}.$ The nonzero primes of $R$ are all maximal and of the form ${\mathfrak m}_i = \{ f \in R \colon f(\alpha + p^i) \in p\mathbb Z_p\}$ ($i\geq0$). The ideal $\mathfrak m_0$ is not finitely generated but all the others are. This follows from general facts about rings of "integer-valued rational functions" together with the observation that $\alpha$ is a limit point of $A$ while $\alpha + p^i$ (with $i>0$) are all isolated points. The relevant details are in Chapter X of Cahen & Chabert, Integer-Valued Polynomials (AMS, 1997); see exercises 19 and 20 in particular.
Incidentally, the localization of $R$ at each $\mathfrak m_i$ is a dvr, but $R$ is not noetherian because ${\mathfrak m}_0$ is not finitely generated. Thus $R$ is an example of an "almost Dedekind domain" that isn't Dedekind. A question about such rings came up not too long ago.
