Let $R$ be a noetherian ring. Then $R$ is Jacobson if and only if for every prime ideal $P$ of $R$ such that $\dim R/P=1$ the ring $R/P$ has infinitely many maximal ideals.
"$\Rightarrow$" Actually any homomorphic image of dimension $\ge 1$ of a Jacobson ring has infinitely many maximal ideals. Since homomorphic images of Jacobson rings are also Jacobson, we can start with a Jacobson ring $R$ of dimension $\ge1$, and prove that $R$ has infinitely many maximal ideals. Suppose the contrary, and let $P$ be a prime ideal of $R$ which is not maximal. Then $P$ is a finite intersection of maximal ideals, and therefore $P$ itself is maximal, a contradiction.
"$\Leftarrow$" If $R$ isn't Jacobson, then there is a prime ideal which is not an intersection of maximal ideals. Since $R$ is noetherian we can choose a prime $P$ maximal with this property. In particular, $P$ isn't maximal, so $\dim R/P\ge 1$.
If $\dim R/P=1$, then, by hypothesis, the intersection of its maximal ideals is $(0)$. (Otherwise, taking a non-zero element $x$ in the intersection, and using the primary decomposition of $(x)$ we find that $x$ is contained in a finite number of maximal ideals, contradiction.) Thus $P$ is an intersection of maximal ideals, and this contradicts the choice of $P$.
If $\dim R/P>1$, then there are infinitely many height one prime ideals in $R/P$ and (by a similar argument to the one in the previous paragraph) their intersection is $(0)$, that is, $P$ is an intersection of larger primes. From the choice of $P$ we deduce that $P$ is an intersection of maximal ideals, contradiction.