Proof of Hilbert Nullstellensatz from Zariski's Lemma On wikipedia there is a proof of Hilbert Nullstellensatz from Zariski's Lemma: link here. The proof is as follows.

Let $A = k[t_1, \ldots, t_n]$ ($k$ algebraically closed field), $I$ an ideal of $A,$ and $V$ the common zeros of $I$ in $k^n$. Clearly, $\sqrt{I} \subseteq I(V)$. Let $f \not\in \sqrt{I}$. Then $f \not\in \mathfrak{p}$ for some prime ideal $\mathfrak{p}\supseteq I$ in $A$. Let $R = (A/\mathfrak{p}) [f^{-1}]$ and $\mathfrak{m}$ a maximal ideal in $R$. By Zariski's lemma, $R/\mathfrak{m}$ is a finite extension of $k$; thus, is $k$ since $k$ is algebraically closed. Let $x_i$ be the images of $t_i$ under the natural map $A \to k$. It follows that $x = (x_1, \ldots, x_n) \in V$ and $f(x) \ne 0$.

There are a few things which I find puzzling. First of all, what is $(A/\mathfrak{p}) [f^{-1}]$? In general for $f \in A,$ $f^{-1}$ might not exist in $A.$ $A$ is not a field.
Then we have the "natural map". The "most" natural map is the following: quotient map $q_1:A \to A/\mathfrak p,$ followed by inclusion $ i:A/\mathfrak p \to (A/\mathfrak{p}) [f^{-1}]=R,$ followed by quotient map $q_2:R \to R/ \mathfrak m \cong k.$ Is this the correct quotient map? If so, how can we guarantee that $i\circ q_1(t_k) \notin \mathfrak m$? No restrictions on the choice of $\mathfrak m$ is stated.
Finally how do we know that $x \in V$?
 A: Question: "There are a few things which I find puzzling. First of all, what is (A/p)[f−1]? In general for f∈A, f−1 might not exist in A. A is not a field."
Answer: If $B$ is a commutative ring and $f\in B$ there is a canonical isomorphism
$$B_f \cong B[t]/(tf-1).$$
Hence in the above argument: Since $A$ is a finitely generated $k$-algebra it follows
$$R:=(A/\mathfrak{p})_f\cong (A/\mathfrak{p})[t]/(tf-1)$$
is a finitely generated $k$-algebra. Hence $R/\mathfrak{m}$ is a finite extension of $k$, and must equal $k$ since $k$ is algebraically closed.
Note: Since $R$  is a finitely generated $k$-algebra it follows $R$ is noetherian hence $R$ has a maximal ideal $\mathfrak{m}$.
Let
$$\phi: A \rightarrow R$$
be the canonical map. The inverse image $\phi^{-1}(\mathfrak{m}):=\mathfrak{n} \subseteq A$ contains the maximal ideal $(t_i-x_i)$ hence $\mathfrak{n}=(t_i-x_i)$. If $g\in I$ it follows
$$\phi(g)=g(x_1,..,x_n)=0$$
hence $p:=(x_1,..,x_n)\in V$. By definition: The element $f/1 \in R$ is a unit, hence $f/1 \notin \mathfrak{m}$, and hence
$$\overline{\phi(f)} \neq 0\text{  in }R/\mathfrak{m}.$$
Hence
$$f(p) \neq 0.$$
