Let $R$ be an arbitrary commutative ring with unit. The nilradical $\mathfrak N = \mathfrak N(R)$ is the intersection of $R$'s prime ideals, hence the closed subset $V(\mathfrak N)$ contains every point of $\operatorname{Spec}(R)$, so, in a sense, $\operatorname{Spec}(R/\mathfrak N)$ is the smallest closed subscheme of $\operatorname{Spec}(R)$ that the raw topology sees as being the whole of $\operatorname{Spec}(R)$, i.e., any distinctions can only be seen in the structure sheaf. I see the inclusion $\operatorname{Spec}(R/\mathfrak N) \to \operatorname{Spec}(R)$ as adding an infinitesimal layer of fat on top of $\operatorname{Spec}(R/\mathfrak N)$, which is just meat and bones. The elements of $R$ and $R/\mathfrak R$ induce the same functions on the same underlying topological space, because we cannot see any nilpotent distinctions by evaluating them at points.
Is there any similar intuition that I could develop for the Jacobson radical? Of course, I can adapt the sentence in the preceding paragraph as follows: $\mathfrak J = \mathfrak J(R)$ is the intersection of $R$'s maximal ideals, hence the closed subset $V(\mathfrak J)$ contains every closed point of $\operatorname{Spec}(R)$, so, in a sense, $\operatorname{Spec}(R/\mathfrak J)$ is the smallest closed subscheme of $\operatorname{Spec}(R)$ that the raw topology sees as containing every closed point of $\operatorname{Spec}(R)$. However, what is a good reason to care about this definition?
Evidently, I am failing to see Nakayama's lemma as a geometric statement.