Let $k$ be an algebraically closed field. Consider a ring $A=k[x_1,\ldots, x_n]$. Our aim is to show that any maximal ideal is of the form $(x_1-x^0_1,\ldots, x_n-x_n^0)$ for some elements $x^0_i \in k$.
To do so, we prove the following fact by the induction on $n$: for any maximal ideal $m$, field $k[x_1,\ldots,x_n]/m$ is isomorphic to $k$.
Proof: case $n=1$, then $k[x_1]$ is PID, hence, $m=(f)$ for some $f=x_1^{k}+\ldots$.
It follows that $k[x]/m$ is finite over $k$, hence, is isomorphic to $k$ (because $k$ is algebraically closed).
Let us now suppose that the claim is true for $n-1$. Let us prove the induction step. Consider the ideal $\tilde{m}:=k[x_1,\ldots, x_{n-1}]\cap m$. It is a maximal ideal of $k[x_1,\ldots,x_{n-1}]$, hence, by induction hypothesis we have $k[x_1,\ldots, x_{n-1}]/\tilde{m}=k$. It follows that $k[x_1,\ldots,x_n]/m=(k[x_1,\ldots,x_{n-1}]/\tilde{m})[x_n]/[m]=k[x_n]/[m]$. Now the desired follows from $n=1$ case.
The Hilbert's Nullstellensatz now directly follows.
The question is the following: this argument seems to be too short to be true (all the proofs I know are much longer), so what is wrong here?