Suppose $K = k[\alpha_1, \ldots, \alpha_m]$ as a $k$-algebra. We want to show all $\alpha_i$ are algebraic over $k$.

Suppose not. After reordering, let $\alpha_1, \ldots, \alpha_r$ be a transcendence basis of $K/k$ (so $r \geq 1$), and $\alpha_{r+1}, \ldots, \alpha_m$ are algebraic over $k(\alpha_1, \ldots, \alpha_r)$.

Then $K$ is a finite extension of $k(\alpha_1, \ldots, \alpha_r) = k(x_1, \ldots, x_r)$ (the rational function field). So $K$ is a finitely generated $k[x_1,\ldots,x_r]$-module. Say $K = k[x_1,\ldots,x_r][\beta_1,\ldots,\beta_s]$ where each $\beta_j$ satisfies a monic polynomial over $k(x_1,\ldots,x_r)$.

Clearing denominators: there exists $0 \neq d \in k[x_1,\ldots,x_r]$ such that each $\beta_j$ is integral over $k[x_1,\ldots,x_r][1/d]$. Hence $K$ is integral over $k[x_1,\ldots,x_r][1/d]$.

But $K$ is a field, so $k[x_1,\ldots,x_r][1/d]$ must also be a field (a subring of a field that is integral over it is a field if the field is integral over it — by the lying-over theorem). But $k[x_1,\ldots,x_r][1/d]$ is a localization of $k[x_1,\ldots,x_r]$, which is not a field when $r \geq 1$ (e.g., the element $x_1 + 1$ is not a unit unless $d$ has infinitely many factors, which is impossible).

Contradiction. So $r = 0$, and all $\alpha_i$ are algebraic over $k$.

Let $\mathfrak{m}$ be a maximal ideal. Then $K = k[x_1,\ldots,x_n]/\mathfrak{m}$ is a field that is a finitely generated $k$-algebra.

By Zariski's Lemma, $K/k$ is algebraic. Since $k = \bar{k}$, we have $K = k$.

The projection $k[x_1,\ldots,x_n] \twoheadrightarrow K = k$ sends $x_i \mapsto a_i$ for some $a_i \in k$. The kernel of this map is $(x_1 - a_1, \ldots, x_n - a_n)$, which is contained in $\mathfrak{m}$. But $(x_1 - a_1, \ldots, x_n - a_n)$ is already maximal, so $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$.

Easy inclusion ($\supseteq$): If $f^r \in \mathfrak{a}$, then for any $P \in V(\mathfrak{a})$, $f(P)^r = 0$ in $k$, so $f(P) = 0$. Thus $f \in I(V(\mathfrak{a}))$.

Hard inclusion ($\subseteq$, the Rabinowitsch trick): Let $f \in I(V(\mathfrak{a}))$. We need to show $f^r \in \mathfrak{a}$ for some $r$.

Introduce a new variable $t$ and consider the ideal

$\mathfrak{b} = \mathfrak{a} + (1 - tf) \subseteq k[x_1, \ldots, x_n, t].$

Claim: $V(\mathfrak{b}) = \varnothing$ in $\mathbb{A}^{n+1}$.

Proof of claim: If $(a_1,\ldots,a_n, b) \in V(\mathfrak{b})$, then $(a_1,\ldots,a_n) \in V(\mathfrak{a})$, so $f(a_1,\ldots,a_n) = 0$ (since $f \in I(V(\mathfrak{a}))$). But also $1 - bf(a_1,\ldots,a_n) = 1 - 0 = 1 \neq 0$. Contradiction. So $V(\mathfrak{b}) = \varnothing$.

By the ideal form of the Nullstellensatz, $\mathfrak{b} = (1)$. So there exist $g_1, \ldots, g_s \in \mathfrak{a}$ and polynomials $h_i, q \in k[x_1,\ldots,x_n,t]$ such that

$1 = h_1 g_1 + \cdots + h_s g_s + q \cdot (1 - tf).$

Now substitute $t = 1/f$ (formally, work in $k[x_1,\ldots,x_n][1/f]$, i.e., localize at $f$):

$1 = h_1(x, 1/f) g_1 + \cdots + h_s(x, 1/f) g_s + 0.$

Multiply both sides by $f^N$ for large enough $N$ to clear all denominators:

$f^N = H_1 g_1 + \cdots + H_s g_s \in \mathfrak{a}$

where $H_i \in k[x_1,\ldots,x_n]$. So $f^N \in \mathfrak{a}$, i.e., $f \in \sqrt{\mathfrak{a}}$.