We let $k$ denote a fixed algebraically closed field of characteristic zero.
We let $R$ denote a reduced finitely generated $k$-algebra where $\dim_kR = n$ as a $k$ vector space and $n$ is a positive integer.
If we write $M$ to be the set of all proper maximal ideals of $R$, how might one show that $|M| = n$? Namely, we are looking to show that $R \cong k^n$ as $k$-algebras through this.
There are several strategies you could pick. Here's one.
First, I claim that $R$ is of Krull dimension zero: we have $k\subset R$ is an integral extension (as for any $0\neq r\in R$ we have that there is some $a>0$ with $r^a\in\operatorname{Span}_k(1,r,r^2,\cdots,r^{a-1})$ by finite-dimensionality) and Krull dimension is preserved by integral extensions. This implies that every prime ideal is maximal, so the intersection of all maximal ideals is the nilradical of $R$; in particular, it is zero, as $R$ is reduced. Hence $R\cong R/0\cong R/\bigcap_{m\subset R} \cong \prod_{m\subset R} R/m$ by the Chinese remainder theorem. Now by Zariski's lemma, the quotient of a finitely-generated $k$-algebra by a maximal ideal is a finite algebraic extension of $k$, which must be $k$ since $k$ is algebraically closed. So $R\cong \prod_{m\subset R} k$, and counting dimensions we have that $R$ has exactly $\dim_k R$ maximal ideals.