$\newcommand{\fm}{\mathfrak{m}}$ Problem 17 in the exercises after the 5th chapter of Atiyah-Macdonald is the following (with some references and hints omitted):
Let $X$ be an affine algebraic variety in $k^n$, where $k$ is an algebraically closed field, and let $I(X)$ be the ideal of $X$ in the polynomial ring $k[t_1,\dots,t_n]$. If $I(X) \ne (1)$ then $X$ is not empty.
Deduce that every maximal ideal in $k[t_1,\dots,t_n]$ is of the form $(t_1-a_1,\dots,t_n-a_n)$ where $a_i \in k$.
The first part follows from the previous exercise, which consists of Noether normalization lemma and proves that $X$ has to surject onto $k^r$ for some $r \ge 0$. But I don't see how to deduce the second part from the first. If we start with a maximal ideal $\fm$, why is it $I(X)$ for some affine algebraic variety $X$? Since $\fm$ is maximal, and is contained in $I(V(\fm))$, it would be enough to show that $I(V(\fm)) \ne (1)$, but that seems to be equivalent to the required conclusion.
I can see how to finish after this, just let $(a_1,\dots,a_n) \in X$, and note that $\mathfrak{m} \subseteq (t_1-a_1,\dots,t_n-a_n)$, by repeated application of the remainder theorem.
$\DeclareMathOperator{\m}{\!\mathfrak{m}}$Showing that $V(\,\m) \ne \emptyset$ (for an algebraically closed field $k$ and a maximal ideal $\,\m \subseteq k[t_1, \ldots, t_n]$) from first principles, is equivalent to the Nullstellensatz! Indeed the subtlety can be seen from the fact that it is no longer true if $k \ne \overline{k}$: e.g. $\,\m = (t^2 + 1) \subseteq \mathbb{R}[t]$ is maximal, but $V(\,\m) = \emptyset$.
Probably what Atiyah-MacDonald means is: use Noether normalization to deduce $\,\m = (t_i - a_i)$. To do this, notice that for any maximal ideal $\,\m$, $k[t_1, \ldots, t_n]/\m$ is a finitely generated $k$-algebra, hence by Noether normalization has a polynomial subring over which it is integral. But this subring must be $k$ itself (i.e. the polynomial ring in $0$ variables), so $k \subseteq k[t_1, \ldots, t_n]/\m$ is an integral extension. Since $k = \overline{k}$ and $k[t_1, \ldots, t_n]/\m$ is a field, we must have equality $k = k[t_1, \ldots, t_n]/\m$. If $\pi : k[t_1, \ldots, t_n] \twoheadrightarrow k[t_1, \ldots, t_n]/\m$ is the canonical projection, setting $a_i := \pi(t_i) \in k$ gives $(t_i - a_i) \subseteq \ker \pi = \m$, and then equality holds since $(t_i - a_i)$ is maximal.