Over which fields $k$ is there a reasonable analogue of Hilbert's Nullstellensatz?

Here is a more precise formulation: let $k$ be an arbitrary field, $n$ a positive integer, and $R = k[t_1,..,t_n]$. There is a natural relation between $k^n$ and $R$: for $x \in k^n$ and $f \in R$, $(x,f)$ lies in the relation if $f(x) = 0$.

This relation induces a Galois connection between the power set of $k^n$ and the set of all ideals of $R$ (both partially ordered by inclusion). In more standard algebraic-geometric language, if $S$ is a subset of $k^n$ and $J$ is an ideal of $R$, put

$I(S) = \{f \in R \ | \ \forall x \in S, \ f(x) = 0\}$

and

$V(J) = \{x \in k^n \ | \ \forall f \in J, \ f(x) = 0\}$.

There are induced closure operators: for a subset $S$, $\overline{S} := V(I(S))$ and for an ideal $J$, $\overline{J} := I(V(S))$.

The closure operator on subsets is compatible with finite unions so is the closure operator for a topology on $k^n$, the Zariski topology.

The question is: what is the closure operator $I \mapsto \overline{I}$ on ideals of $R$? By a Nullstellensatz, I mean a nice description of this closure operator.

Some remarks and examples:

1) Over any field $k$, one sees that $\overline{I}$ is a radical ideal hence contains $\operatorname{rad}(I) = \{x \in R \ | \ \exists n \in \mathbb{Z}^+ \ | \ x^n \in I\}$.

If $k$ is algebraically closed, then Hilbert's Nullstellensatz says that $\overline{I} = \operatorname{rad}(I)$.

It is easy to see that if $\overline{I} = \operatorname{rad}(I)$ for all maximal ideals of $k[t]$, then $k$ is algebraically closed.

2) If $k$ is formally real, then for any ideal $I$ of $R$, $\overline{I}$ is a **real ideal**, i.e., $x_1,\ldots,x_n \in R, \ x_1^2 + \ldots + x_n^2 \in I \implies x_1,\ldots,x_n \in I$. Moreover, for any ideal $I$ in a commutative ring, there is a unique minimal real ideal containing $I$, its **real radical** $\mathbb{R}ad(I)$, which is the intersection of all real prime ideals $\mathfrak{p}$ containing $I$.

If $k$ is real-closed, then for any ideal $I$ in $k[t_1,\ldots,t_n]$, $\overline{I} = \mathbb{R}ad(I)$: this is Risler's Nullstellensatz.

3) There is also a Nullstellensatz for p-adically closed fields (in particular, for $p$-adic fields) due to Jarden and Roquette: see

Are there further Nullstellensätze (say, for non-Henselian fields to rule out variations on 3)?

Although I haven't been precise on what a description of $\overline{I}$ means (I don't know how), it seems reasonable to guess that there is no good Nullstellensatz over a field like $\mathbb{Q}$ for which it is believed that Hilbert's 10th problem has a negative answer. Briefly: if you had a system of polynomial equations $P_1,\ldots,P_m$ with $\mathbb{Q}$-coefficients, then they have a simultaneous solution over $\mathbb{Q}$ iff the closure of $\langle P_1,\ldots,P_m \rangle$ is a proper ideal, so if you had a sufficiently nice description of the closure operation, you could use it to answer H10 over $\mathbb{Q}$ affirmatively.

A case of persistent interest to me over recent years is that of a finite field. In some sense this is the worst case, since it is not hard to show that the zero ideal in $k[t_1,\ldots,t_n]$ is closed iff $k$ is infinite. Nevertheless, I vaguely feel like there should be something to say here, possibly something having to do with reduced polynomials -- i.e., for which each exponent of each variable is at most $\# k - 1$ -- as in one of the proofs of the Chevalley-Warning theorem.

P.S.: I am aware of other algebraic results about $k[x_1,\ldots,x_n]$ over a general field $k$ which, when $k$ is algebraically closed, imply Hilbert's Nullstellensatz, e.g. that a finitely generated $k$-algebra which is a field is finite-dimensional over $k$, or that every prime ideal in $k[t_1,\ldots,t_n]$ is an intersection of maximal ideals. These are interesting and useful, but here I am really interested in $I \mapsto \overline{I}$.