Assume that $I$ is a proper radical ideal of $k[x_1,...,x_n]$ which is not a maximal ideal where $k$ is a field. Prove that $I$ is contained in at least two maximal ideals.
Now it is easy to prove the claim using Hilbert's Nullstellensatz. I wonder if there is a more elementary argument to prove this claim.
I do not know if it is sufficiently elementary, but you can proof this using Noether normalization and Going Up.
Let $I$ be radical ideal, such that $R=k[x_1, \dotsc, x_n]/I$ is a local ring. We have to show that $R$ is a field.
By Noether normalization, we have a finite map
$$k[y_1, \dotsc, y_d] \hookrightarrow R$$
Since this map is integral, we know that an ideal in $R$ is maximal if and only if its contraction is. In particular - using that $R$ is local - there is only one maximal ideal of $k[y_1, \dotsc, y_d]$, which is a contraction of a prime of $R$.
Let $\mathfrak p \subset R$ be a minimal prime ideal and $\mathfrak q$ be the contraction along this map. By Going Up, all maximal ideals of $k[y_1, \dotsc, y_d]$ containing $\mathfrak q$ are obtained as a contraction from a prime of $R$. By the argument above, there is only one such maximal ideal. Thus we have shown, that $\mathfrak q \subset k[y_1, \dotsc, y_d]$ is a radical ideal, which is contained in a unique maximal ideal. Using induction (note that $d < n$), we get that $\mathfrak q$ is maximal, i.e. $\mathfrak p$ was maximal.
This shows that $R$ is a zero-dimensional local ring with maximal ideal $\mathfrak p = \operatorname{nil} R$. It is also reduced, hence $\mathfrak p=0$ and thus $R$ is a field.
Of course you can use similiar arguments to obtain the Nullstellensatz from Noether normalization and Going Up.
I am pretty sure that basically every proof of this statement will come very close to proving the Nullstellensatz as well.
