In artin's Algebra book (1st edition) the following theorems are stated:
(Hilbert's Nullstellensatz) In $\mathbb{C}[x_1,…,x_n]$ maximal ideals are of the form $\langle x_1-a_1,…,x_n-a_n\rangle$.
(Classical Nullstellensatz) $f_1,…,f_n\in \mathbb{C}[x_1,…,x_n] $. If $g=0$ in the variety defined by zeros of $f_1,…,f_n$, then there is a power $g^m\in \langle f_1,…,f_n\rangle$
He then asks us to prove that the Hilbert Nullstellensatz is a consequence of the Classical Nullstellensatz.
I think I was able to prove this, but it is not exactly pretty, I am looking for nicer solutions and for any possible flaw in my argument. Take $\mathcal{M}$ a maximal ideal and suppose $\mathcal{M}\not= \langle x_1-a_1,…,x_n-a_n\rangle$ for any $(a_1,…,a_n)\in \mathbb{C}^n$. Because $\mathcal{M}$ is maximal and $\langle x_1-a_1,…,x_n-a_n\rangle$ are proper ideals of our ring this means that:
$$\mathcal{M}\subset \langle x_1-a_1,…,x_n-a_n\rangle \quad \text{cannot hold.} $$
In other words, for every $(a_1,…,a_n)\in \mathbb{C}^n$ there is $m_{(a_1,…,a_n)}\in \mathcal{M}$ but not in $ \langle x_1-a_1,…,x_n-a_n\rangle $. This means that $m_{(a_1,…,a_n)}(a_1,…,a_n)\not=0$. Similarly, because $\langle x-a_1,…,x-a_n\rangle$ are maximal ideals, if it were contained in $\mathcal{M}$ it would be $\mathcal{M}$, so we also need $m^{(a_1,…,a_n)}\in \langle x-a_1,…x-a_n\rangle$ but not in $ \mathcal{M}$.
Consider the following variety where $f_1,…,f_k$ are drawn from $\mathcal{M}$, which has $d\geq 1$ zeros given by $k_1,…,k_d\in \mathbb{C^n}$.
$$\begin{cases}
f_1=0\\
….\\
f_k=0\end{cases}$$
$m^{k_1}\cdot…\cdot m^{k_d}$ is zero in the variety, so $(m^{k_1}\cdot…\cdot m^{k_d})^n\in \langle f_1,…,f_k\rangle\subset \mathcal{M}$ by the Nullstellensatz. Because $\mathcal{M}$ is maximal it is prime which implies that $m^{k_1}\cdot…\cdot m^{k_d}\in \mathcal{M}$ and $m^{k_j}\in \mathcal{M}$ for some $j$ which is absurd!
Hence a variety drawn from $k$ polynomials of $\mathcal{M}$ has no zeros. But this implies by a proposition from Artin:
$$g_1f_1+…+g_k f_k=1\in \mathcal{M}$$
So $\mathcal{M}=\mathbb{C}[x_1,…,x_n]$ which is absurd.
Best Answer
First we record a corollary of the classical theorem. (The proof is behind the spoiler block, in case you want to prove it yourself.)
Corollary: If $I, J \subset \mathbb{C}[x_1, \dots, x_n]$ are ideals where $Z(I) = Z(J)$ as subsets of $\mathbb{C}^n$, then $\sqrt{I} = \sqrt{J}$.
Now, if $\mathcal{M} \subset \mathbb{C}[x_1, \dots, x_n]$ is a maximal ideal. Then $Z(\mathcal{M}) \subset \mathbb{C}^n$ is nonempty by the corollary, so we can find some $(a_1, \dots, a_n) \in Z(\mathcal{M})$. It then follows that $\mathcal{M} \subseteq (x_1 - a_1, \dots, x_n - a_n)$, which implies the desired result.
EDIT: Let $I \subset \mathbb{C}[x_1, \dots, x_n]$ be an ideal. Then we define the zero locus of $I$ to be $$Z(I) = \{a \in \mathbb{C}^n \;|\; f(a) = 0\text{ for all }f\text{ in } I\}$$ and the radical of $I$ to be $$\sqrt{I} = \{f \in \mathbb{C}[x_1, \dots, x_n]\;|\; f^n \in I \text{ for some } n \geq 0\}.$$