I'm addressing people familiar with the vernacular of logicians.(The tag commutative algebra is due anyway.)
TL;DR The following atypical proof of Hilbert's Basis Theorem puzzle me. The reason, at the end.
Satz 1. Let $A\subseteq M\models T_{\rm id}$ and let $p(x)\subseteq L_{\rm at^\pm}(A)$, where $x$ is a finite tuple. Then there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p_+(x)$, where $p_+(x)=p(x)\cap L_{\rm at}(A)$.
Notation
$T_{\rm id}\ $ theory of integral domains in the language of rings
$L_{\rm at}(A)\ $ atomic formulas (or, up to equivalence, polynomial equations)
$L_{\rm at^\pm}(A)\ $ atomic and negated atomic formulas
$\vdash\ $ logic consequence over $T_{\rm id}\cup{\rm Diag}\langle A\rangle_M$
Motivation
Only to provide some context. Hilbert's Basis Theorem translated in a model theoretical language claims the following
Satz 2. Let $A\subseteq M\models T_{\rm id}$ and let $p(x)\subseteq L_{\rm at}(A)$, where $x$ is a finite tuple. Then there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p(x)$.
Edit 1: when $A=M$ is a field this is the translation of $M[x]$ is Noetherian, a weaker form of Hilbert's Basis Theorem, enough for my purposes.
Edit 2: Edit 1 is not completely correct. As remarked by Alex Kruckman, the definition of $\vdash$ above, entails that types closed under $\vdash$ correspond to radical ideal. Hence Satz 2 only claims that in $M[x]$ radical ideal are finitely generated.
Obviously Satz 2 $\Rightarrow$ Satz 1. The converse implication follows from a standard compactness argument. Parenthetically, Hilbert's Nullstellensatz easily follows from Satz 1, by quantifier elimination.
Proof of Satz 1
By induction on the length of $x$.
If $x$ is the empty tuple, the claim is trivial.
Now, let $x$ be finite tuple and let $y$ be a single variable. Let $p(x,y)\subseteq L_{\rm at^\pm}(A)$. We write $p(x)$ for the set of formulas $\varphi(x)\in L_{\rm at^\pm}(A)$ such that $p(x,y)\vdash\varphi(x)$.
Assume as induction hypothesis that there is a conjunction of formulas in $p(x)$, say $\psi(x)$, such that $\psi(x)\vdash p_+(x)$. We prove Sats 1 with $x,y$ for $x$.
Let $\psi(x,y)\in L_{\rm at}(A)$ have minimal degree (in $y$) among the formulas such that $p(x,y)\vdash\psi(x,y)$ and $\psi(x)\nvdash\psi(x,y)$. Let $n>0$ be the degree of $\psi(x,y)$, so we may assume $\psi(x,y)$ has the form $t(x,y)=0$ where
$$t(x,y)=t_n(x)\cdot y^n + t'(x,y),$$
$t'(x,y)$ is a polynomial of degree $<n$, and $p(x)\nvdash t_n(x)=0$. We claim that
$$(1)\quad p_+(x,y)\dashv t_n(x)\neq0\; \wedge\; \psi(x)\;\wedge\; \psi(x,y)$$
Suppose not and let $s(x,y){=}0\in p_+(x,y)$ be a counter example to 1 of minimal degree. By the minimaly of $n$ we may assume that $s(x,y)$ has the form
$$s(x,y)=s_{n+i}(x)\cdot y^{n+i}+s'(x,y)$$
for some polynomial $s'(x,y)$ of degree $<n+i$ and some $i\ge0$. Clearly $$(2)\quad p(x,y)\vdash s_{n+i}(x)\cdot t(x,y)\cdot y^{i} - t_n(x)\cdot s(x,y) = 0.$$
It is also clear that the polynomial in (2) has degree $<n+i$. This contradicts the choice of $s(x,y)$. A contradiction that proves the proposition.
Question
Note that an elementary argument proves Satz 1 for tuples of length 1. Therefore I would expect the induction to go as follows. Let $a$ realize $p(x)$ in some large model. Apply to $p(a,y)$ the elementary for tuples of length 1 and obtain that $\psi(a,y)\vdash p_+(a,y)$ for some conjunction of formulas in $p(a,y)$. Conclude that Satz 1 holds for $p(x,y)$.
The strategy above does not work. Hence I suspect that a stronger claim than Satz 1 holds (that would make the strategy work).