Assume all rings/algebras are commutative with identity. "$\hookrightarrow$" means injection and "$\twoheadrightarrow$" means surjection. $A_f$ is the localization of $A$ on the point $f$. We slightly abuse $\mathfrak p\cap R$ to denote the retraction $f^{-1}(\mathfrak p)$ of the prime ideal $\mathfrak p\subset S$ in the ring map $f:R\to S$.
$A$ is a Jacobson ring if every prime ideal is the intersection of the maximal ideals containing it.
The word "Simple" refers to that we can apply the standard techniques (see "Proof of $(\color{RoyalBlue}2)$") used to prove the Nullstellensatz for fields. The key is to prove $\operatorname{Im} (A/\mathfrak p\to (A/\mathfrak p)_f/\mathfrak m)$ is a field and construct a maximal ideal as $\ker (A/\mathfrak p\to (A/\mathfrak p)_f/\mathfrak m)$ in $A/\mathfrak p$ in the arrow compositions below: $$k\to A/\mathfrak p\hookrightarrow (A/\mathfrak p)_f\twoheadrightarrow (A/\mathfrak p)_f/\mathfrak m$$
Here is the form of Nullstellensatz I would like to prove
If $R$ is a Jacobson ring, then so is any finite generated $R$-algebra $A$.
Here is my proof: Assume $A$ is not Jacobson and choose a prime ideal $\mathfrak p\subset A$ and an element $f$ such that $f\notin \mathfrak p$ but $f$ is contained in all maximal ideals containing $\mathfrak p$. We have composition arrows $$R/(\mathfrak p\cap R)\to A/\mathfrak p\hookrightarrow (A/\mathfrak p)_f\twoheadrightarrow(A/\mathfrak p)_f/\mathfrak m_{\mathfrak p}$$ where $\mathfrak m_{\mathfrak p}$ is a maximal ideal of $(A/\mathfrak p)_f$. Inspired by the classic Nullstellensatz for fields, we will need $\mathfrak p\cap R$ to be large enough to make $R/(\mathfrak p\cap R)$ a field. To be precise, we make a hypothesis and assume it is true.
Hypothesis. There is a prime ideal $\mathfrak q\supset \mathfrak p$ such that $f\notin \mathfrak q$ and $\mathfrak q\cap R$ is maximal in $R$.
Then looking at the composition arrows again for $\mathfrak q$: $$R/(\mathfrak q\cap R)\to A/\mathfrak q\hookrightarrow (A/\mathfrak q)_f\twoheadrightarrow(A/\mathfrak q)_f/\mathfrak m_{\mathfrak q}$$
$(A/\mathfrak q)_f/\mathfrak m$ is a finite field extension of the field $R/(\mathfrak q\cap R)$. As an intermediate extension, the image of $A/\mathfrak q$ in $(A/\mathfrak q)_f/\mathfrak m$ is also finite over $R/(\mathfrak q\cap R)$, and must then be a field. Thus $\ker(A/\mathfrak q\to (A/\mathfrak q)_f/\mathfrak m)$ is a maximal ideal in $A/\mathfrak q$, and corresponds to a maximal ideal in $A$ containing $\mathfrak p$ but not containing $f$, contradiction.
The hypothesis is true, for example this gives a sharper result, showing $(A/\mathfrak q)_f$ is a field. And then use the fact that $R/(\mathfrak q\cap R)\hookrightarrow (A/\mathfrak q)_f$ is a finite type extension from a Jacobson ring to a field to deduce $R/(\mathfrak q\cap R)$ is a field, as shown here.
Is there a easier way through to prove the hypothesis above without showing $(A/\mathfrak q)_f$ is a field, but rather studying the prime/maximal correspondence between $R\to A$?
Update: I am looking at a particularly interesting result which will prove our hypothesis:
If $R$ is Jacobson, then for any finite type ring map $f: R\to A$, $f^{-1}(\mathfrak m)$ is maximal in $R$ if $\mathfrak m$ is maximal in $A$.
To prove our hypothesis using this result, form the composition ring map $R\to A\to (A/\mathfrak p)_f$, choose a maximal ideal of $(A/\mathfrak p)_f$ and look at the corresponding prime ideal in $A$ and in $R$.
Update: The converse of the result is also true, and strongly relies on this result. I believe that is why stacks project uses the sharper result to prove the Nullstellensatz for Jacobson rings.
$R$ is Jacobson if and only if for any finite type ring map $R\to A$, inverse image of maximal ideals are maximal as well.