Let $k$ be a field and let $K$ be an extension of $k$ of degree larger than $1$. Let $R=k[X_i, i\in K]$ be a polynomial ring with one variable per element of $K$. There is a $k$-linear ring homomorphism $\phi:R\to K$ which maps $X_i$ to $i$ for all $i\in K$, and it is surjective. It follows that the kernel of $\phi$ is maximal. Clearly, it is not of the form you mentioned: indeed, the quotient of $R$ by all ideals of that form is $1$-dimensional as a $k$-algebra.
Let $k$ be a field and $\bar{k}$ its algebraic closure. An algebraic zero of a subset $\Phi$ of $k[x_1,\dots,x_n]$ is an element $(a_1,\dots,a_n) \in \bar{k}^n$ such that $f(a_1,\dots,a_n)=0, \, \forall f \in \Phi$. Then Hilbert's Nullstellensatz says that if $g \in k[x_1,\dots,x_n]$ vanishes at every algebraic zero of $\Phi$, then $g$ is inside the radical of the ideal generated by $\Phi$ (Matsumura, Theorem 5.4).
The next key thing to observe is that given an ideal $I$ of $k[x_1,\dots,x_n]$, there is a $1-1$ correspondence between algebraic zeros of $I$ and maximal ideals of $k[x_1,\dots,x_n]$ that contain $I$. To see that, note that if $m$ is a maximal ideal that contains $I$ and we define $a_i$ to be the class of $x_i$ mod $m$, then $(a_1,\dots,a_n)$ is an algebraic zero of $I$ by the Zariski Lemma. Conversely, if $(a_1,\dots,a_n)$ is an algebraic zero, then $k[a_1,\dots,a_n] = k(a_1,\dots,a_n)$ and the kernel of the $k$-algebra homomorphism $k[x_1,\dots,x_n] \rightarrow k[a_1,\dots,a_n]$ that sends $x_i$ to $a_i$ is a maximal ideal (Matsumura, Theorem 5.1).
Finally, we clearly have that $I \subset \cap_{m \supset I} m$. Conversely, let $f \in \cap_{m \supset I} m$. Then $f$ vanishes at every algebraic zero of $I$ and by Hilbert's Nullstellensatz $f \in \sqrt{I}$.
Best Answer
The inclusion $\displaystyle \sqrt{I} \subseteq \bigcap_{m \supseteq I \text{ maximal}} m$ always holds. Suppose $\displaystyle f \in \bigcap_{m \supseteq I \text{ maximal}} m$. Since $k$ is algebraically closed, the maximal ideals containing $I$ are precisely the ideals of the form $(x_1 - a_1, \ldots, x_n - a_n)$, where $(a_1, \ldots, a_n)$ are the points of $V(I) \subseteq k^n$. This says that $f$ vanishes at every point of $V(I)$, so $f \in I(V(I)) = \sqrt{I}$, by the Nullstellensatz.
This in fact still holds even if $k$ is not algebraically closed, and says that $k[x_1, \ldots, x_n]$ (for any field $k$) is a Jacobson ring.