This turns out to be true for any completely regular space $X$:
Let $X$ be completely regular having the property that every maximal ideal in $C(X)$ is of the form $M_x$. To show that $X$ is compact, let $\mathcal C$ be a family of closed sets having the finite intersection property. We need to prove that $\bigcap_{C\in \mathcal C} C$ is nonempty.
Let $I \subset C(X)$ be the set of continuous functions such that $f^{-1}(0) \supset C$ for some $C\in \mathcal C$. Then - by complete regularity** - we have
$\bigcap_{f\in I} f^{-1}(0) = \bigcap_{C\in \mathcal C} C$
and $I$ is an ideal in $C(X)$ as is easily checked. But then $I$ is contained in some maximal ideal $J$, which by assumption is of the form $M_x$ for some $x\in X$. So we must have $x \in \bigcap_{f\in I} f^{-1}(0) = \bigcap_{C\in \mathcal C} C$ and in particular $\bigcap_{C\in \mathcal C} C \ne \emptyset$.
Add.: ** If $X$ is completely regular, $C\subset X$ is closed and $x\notin C$, then there exists a continuous function $f$ with $f(x) = 1$, $f(C) \subset \{0\}$. This implies that $x\notin \bigcap_{f: f^{-1}(0) \supset C} f^{-1}(0)$ in this case. Hence $X\setminus C$ does not intersect $\bigcap_{f: f^{-1}(0) \supset C} f^{-1}(0)$ and we must have
$C = \bigcap_{f: f^{-1}(0) \supset C} f^{-1}(0)$
Now take the intersection over all $C \in \mathcal C$.