Your proof looks okay.
Here is a reason why it is called the Nullstellensatz:
Assume that your ground field $k$ is algebraically closed. Let $B = k[x_1,...,x_n]$ where the $x_i$ are indeterminates. Let $m \subset B$ be a maximal ideal. Then $B/m$ is a field extension of $k$ which is also clearly a finitely generated $k$ algebra. Then by what you proved above, $k \hookrightarrow B/m$ is a finite ring map. However, as $k$ is algebraically closed, this means that the ring map must be an isomorphism. Let $a_i \in k$ be those elements that are mapped to $x_i + m$ in $B/m$. Then it follows that $x_i - a_i \in m$, which implies that $(x_1 - a_1, ..., x_n - a_n) \subset m$. But, $(x_1 - a_1, ..., x_n - a_n)$ is a maximal ideal, so we must have $(x_1 - a_1, ..., x_n - a_n) = m$. What we have just proved is usually referred to as Hilbert's Weak Nullstellensatz (or so I think).
You can use the Weak Nullstellensatz to prove the Strong Nullstellensatz (a reference would be Mumford's Red Book), which says that following:
For an algebraically closed field $k$, given an ideal $\alpha \subset k[x_1,...,x_n]$, we have $I(V(\alpha)) = \sqrt{\alpha}$, where
\begin{equation}
V(\alpha) = \{\vec{a} \in k^n: \forall f \in \alpha, f(\vec{a}) = 0 \}
\end{equation}
and
\begin{equation}
I(Z) = \{f \in k[x_1,...,x_n]:\forall \vec{a} \in Z, f(\vec{a}) = 0\}
\end{equation} for $Z \subset k^n$.
You can now see what this has to do with the zeroes of polynomials. As a side note, the Strong Nullstellensatz also has an equivalent algebraic formulation that amounts to proving that $k[x_1,...,x_n]$ is a Jacobson ring, that is a ring where any radical ideal is the intersection of the maximal ideals containing it. Hope this helps, and let me know if you are confused by anything.
$\textbf{Later edit:}$ I should mention that the Weak Nullstellensatz is also a very geometric statement. It clearly implies that the elements of any proper ideal of $k[x_1,...,x_n]$ have a common zero.
Also, while I like the proof of the statement that I refer to as the Strong Nullstellensatz using Rabinowitsch's Trick, I find it difficult to remember the trick. Here is an alternate proof to the algebraic version of the Strong Nullstellensatz using techniques that I am sure you are familiar with, since you are reading chapter $5$ of A-M.
$\textbf{The Nullstellensatz:}$ Let $k$ be any field, not necessarily algebraically closed, and let $A$ be a nonzero finitely generated $k$ algebra. Then we have the following:
(a) $\sqrt{(0)} \subset A$ is the intersection of all maximal ideals containing it.
(b) Given an ideal $I \subset A$, $\sqrt{I}$ is the intersection of all maximal ideals containing it.
$\textbf{Proof:}$ (a) I will let $M$ denote the set of maximal ideals of $k[x_1,...,x_n]$. Since $\sqrt{(0)}$ is the intersection of all prime ideals containing it, it follows that $\sqrt{(0)} \subset \cap_{m \in M} m$. Let $f$ be an element contained in every maximal ideal, and suppose that $f \notin \sqrt{(0)}$. Then $A_f$ is a nonzero ring, hence contains a maximal ideal $m$. Note that $A_f$ is also a finitely generated $k$ algebra (why?), and as a result, so is $A_f/m$. By Zariski's Lemma, this means that $A_f/m$ is a finite extension of $k$. Let $p: A \rightarrow A_f/m$ be the standard map. Then we have the inclusions $k \subset A/p^{-1}(m) \subset A_f/m$. Since $A_f/m$ is integral over $k$, it follows that $A/p^{-1}(m)$ is also integral over $k$. But $A/p^{-1}(m)$ is a domain, so it must be a field. It follows that $p^{-1}(m)$ is a maximal ideal of $A$, and by construction $f \notin p^{-1}(m)$, which contradicts our choice of $f$. This proves (a).
(b) follows from (a) by passing to the quotient $A/I$ (do you see why?). $\blacksquare$
As an exercise try to show that $I(V(\alpha)) = \sqrt{\alpha}$ over an algebraically closed field $k$, using the above theorem.
Best Answer
$A[u^{-1}]$ is naturally isomorphic to the localization of $A$ by the multiplicative set $S=\{1,u,u^2,u^3,...\}$. Indeed, any element of $A[u^{-1}]$ can be written in the form $\frac{a}{u^n}=au^{-n}$ for some $a\in A$ and $n\geq 0$, and two such representations are equal exactly when they are equal in $S^{-1}A$.
So now the existence of the extension just follows from the universal property of localization, as the assumption is that $u$ (and so all the elements of $S$) is mapped to an invertible element of $\Omega$.