It's hard for me to think of an area of algebra that applied model theorists haven't touched recently. I have not heard of any logicians working on Iwasawa theory, but it wouldn't surprise me if there are some.
Diophantine geometry: here is a survey article by Thomas Scanlon on applications of model theory to geometry, including discussions of Mordell-Lang and the postivie-characteristic Manin-Mumford conjecture.
Number fields: Bjorn Poonen has shown that there is a first-order sentence in the language of rings which is true in all finitely-generated fields of characteristic 0 but false in all fields of positive characteristic. It was conjectured by Pop that any two nonisomorphic finitely-generated fields have different first-order theories.
Polynomial dynamics: see here for a recent preprint by Scanlon and Alice Medvedev. It turns out that first-order theories of algebraically closed difference fields where the automorphism is "generic" are quite nice.
Differential algebra: By some abstract model-theoretic nonsense ("uniqueness of prime models in omega-stable theories"), it follows that any differential field has a "differential closure" (in analogy to algebraic closure) which is unique up to isomorphism over the base field. There are much more advanced applications, e.g. here.
Geometric group theory: Zlil Sela has recently shown that any two finitely-generated nonabelian free groups are elementarily equivalent (i.e. they have the same first-order theory). According to the wikipedia article, this work is related to his solution of the isomorphism problem for torsion-free hyperbolic groups, but I don't understand this enough to say whether this counts as an "application" of model theory.
Exponential fields: Boris Zilber has suggested a model-theoretic approach to attacking Schanuel's Conjecture. His conjecture that the complex numbers form a "pseudo-exponential field" is actually a strengthening of Schanuel's Conjecture, but the picture that it suggests is appealing. See here for more.
This is in addition to the work on Tannakian formalism, valued fields, and motivic integration that have already been mentioned in other answers, and I haven't even gotten to all the work by the model theorists studying o-minimality. This was just a pseudo-random list I've come up with spontaneously, and no offense is meant to the areas of applied model theory that I've left off of here!
This is to answer the last question on why the $F_i$ form a regular sequence.
In my first attempt I was trying to do this without using the unmixedness of an ideal $I$ generated by $\mathrm{height } I$ number of elements, but it seems that I cannot. (Hat tip to Hailong).
So here is the statement that one needs:
Thm Let $I\subseteq A$ be an ideal generated by $\mathrm{height } I$ number of elements. If $A$ is Cohen-Macaulay, then $I$ is unmixed, that is, it does not have any embedded primes. In other words, every associated prime of $I$ is a minimal prime.
I will certainly not include a proof, because there is no point repeating one of the proofs in the literature. A relatively simple one can be found in Bruns-Herzog's Cohen-Macaulay rings (Cambride University Press). I think if you are willing to discuss regular sequences, then this should be OK and you may have already discussed the Cohen-Macaulay property, or even this theorem, although I kind of doubt it.
However, you don't actually need to discuss CM for this. In fact, this theorem is where the name "Cohen-Macaulay ring" comes from, because it is true that if every such ideal is unmixed, then the ring is Cohen-Macaulay and more importantly, the start of the CM property is a theorem of Macaulay from 1916 that states that the above theorem holds if $A$ is a polynomial ring. Then the next move towards creating this famous name was a theorem of Cohen from 1946 that states that the above theorem holds if $A$ is a regular local ring. Clearly you only need Macaulay's theorem and there must be a simple proof of that somewhere. Perhaps you can adopt the proof from Bruns-Herzog for that special case and make it shorter. There is also a reference to Macaulay's original paper in that book, but I doubt that that's a really good solution to go back to, but you can certainly try.
So, assuming this theorem here is how to prove that
Proposition If $f_1,\dots, f_n$ are polynomials in $n$ variables such that $Z(f_1,\dots,f_n)$ is finite, then $f_1,\dots, f_n$ is a regular sequence.
(this is not exactly how you stated it, but you can get to this situation easily by choosing a hyperplane that misses all the intersection points and restrict to the complement)
Claim 1
If $Z(f_1,\dots,f_n)$ is finite, then any irreducible component of $Z(f_1,\dots, f_r)$ is of dimension $n-r$ for any $1\leq r\leq n$.
Proof Let $W$ be an irreducible component of $Z(f_1,\dots, f_r)$. Then by Krull's principal ideal theorem $\dim W\geq n-r$ and
$\dim (W\cap Z(f_{r+1},\dots, f_n))\geq \dim W -(n-r)$. Since by assumption $\dim (W\cap Z(f_{r+1},\dots, f_n))=0$, it follows that $\dim W\leq n-r$ and hence has to be equal to it. $\square $
Corollary The ideal generated by $(f_1,\dots, f_r)$ has $\mathrm{height}=r$ and hence by the Thm it is unmixed.
Claim 2: Let $M$ be a finitely generated module over the ring $A$ and $x\in A$. Assume that all associated primes of $M$ have the same height and that $\dim M/xM<\dim M$. Then $x$ is not a zero-divisor on $M$.
Proof: The assumption implies that $x$ cannot be contained in any associated prime of $M$ and hence cannot be a zero divisor. $\square $
Claim 1, Corollary, and Claim 2 combined imply that $f_1,\dots,f_n$ is a regular sequence.
Best Answer
To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \mathrm{Spec}(R)\to\mathrm{Spec}(\mathbb Z)$ is of finite type, a theorem of Chevalley says that the image of any constructible subset is constructible. So the image of any closed point by $f$ is a point which is a constructible subset. This can not be the generic point of $\mathrm{Spec}(\mathbb Z)$, so it must be a closed point.
Note that this does not hold in general. For example, over the ring of $p$-adic integers, the ideal $(pX-1)\mathbb{Z}_p[X]$ is maximal, but its preimage in $\mathbb{Z}_p$ is $0$ and it not maximal.
[EDIT] Another proof using Noether's normalization lemma: Noether's normalization lemma over a ring A: if a maximal ideal $\mathfrak m$ of $R$ is such that $\mathfrak m\cap \mathbb Z=0$, then $R/\mathfrak m$ is finite type over (and contains) $\mathbb Z$. So there exits $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z_f[X_1,\dots, X_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z_f[X_1,\dots, X_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers.