The affine Nullstellensatz tells us that we have an inclusion-reversing bijection between radical ideals of $A=k[x_1,\ldots,x_n]$ and affine varieties of $\mathbb{A}^n$, given by $\mathbb{V}\colon A\to\mathbb{A}^n$ and $\mathbb{I}\colon\mathbb{A}^n\to A$.
A fact that is often mentioned is that, under this correspondence, maximal ideals correspond to points in affine space.
Proofs for this often look at the kernel of the evaluation map, and use the fact that $k$ is algebraically closed along with some ring theory, and are not just one-line proofs.
Here is a sketch proof of the above fact; I am convinced that it is wrong, but can't see why:
Let $m$ be a proper maximal ideal of $A$. Now
$X=\mathbb{V}(I)\subseteq\mathbb{V}(m)$ if and only if
$m\subseteq I$. But since $m$ is maximal, either $I=A$ or
$I=m$. So $\mathbb{V}(m)$ has no non-trivial subvarieties apart from
itself and $\emptyset$. But any $a\in\mathbb{V}(m)$ is a subvariety
(since $a=\mathbb{V}(x-a)$). So either $\mathbb{V}(m)=\emptyset$,
which cannot happen since $k$ is algebraically closed, or
$\mathbb{V}(m)=\{a\}$.
Edit:
Thinking about it for a second, the fact that $k$ is algebraically closed does not give us that $\mathbb{V}(m)$ is non-empty straight away, it needs some extra work.
Best Answer
Okay so I'll just dive in here. What is the logic supposed to be? As written the proof makes almost no sense to me, and while it is 4 in the morning here I think there are some serious errors. You have a variety $X = \mathbb{V}(I)$ and you start by assuming $\mathbb{V}(I) \subseteq \mathbb{V}(m)$ which is only true if $m \subseteq I$ which is only true if $I$ is not proper or if $I = m$, so you have shown that $\mathbb{V}(m)$ has no non-trivial subvarieties.
This is good, and of course this is true, but this is just the inclusion reversing correspondence between ideals and closed sets in the corresponding affine space. Note that you have not shown that $\mathbb{V}(I)$ has no non-trivial subvarieties, which is what you wrote (the $X$ should be on the other side of the inclusion). Now comes the really strange part of your proof, why did you not just start at the end? I.e. you know that $\mathbb{V}(m)$ is nonempty by what exactly...? In some sense this is precisely the nullstellensatz (at least in its weak formulation)! The nullstellensatz also gives you exactly what you are trying to prove, note that algebraic closure of the base field gives you absolutely nothing here beyond that it is required to prove the nullstellensatz.
So I am confused as to where you are starting here, you should try to state which theorems you are using. In addition you did not even attempt a proof of the opposite direction: that each point corresponds to a maximal ideal, so your proof is technically incomplete for that reason, though this is a minor quibble.
Edit: so apparently I was confused about the author's predicates in this question. With the nullstellensatz the proof given above is more or less valid, along with almost any other proof of this statement.