I realize that this is an old question, but it was never answered here. So I'll take a crack at it: no, not every ideal is of the form $\frak{m}\!$$_Z$.

Let $M$ be a compact Hausdorff manifold. Then (1) every ideal is contained in some $\frak{m}\!$$_x$, and (2) there are ideals strictly contained in a unique $\frak{m}\!$$_x$.

$\textbf{Proof (1):}$ Let $I$ be a proper ideal of $\mathcal{C}$, the smooth $\mathbb{R}$ valued functions on $M$. Define $\mathcal{Z}[f]:=\{x\in M|f(x)=0\}$, and suppose that $\bigcap_{f\in I}\mathcal{Z}[f]=\emptyset$. Then $M=\bigcup_{f\in I}\mathcal{Z}[f]^c$ and because $M$ is compact there is some finite collection of functions $\{f_i\}_{i=1}^n$ such that $M=\bigcup_{i=1}^n\mathcal{Z}[f_i]^c$. But that would mean that $\bigcap_{i=1}^n\mathcal{Z}[f_i]=\emptyset$, so $\sum_{i=1}^nf^2_i\in I$ would be a unit, so $I=\mathcal{C}$. Then it must be that, $\exists x\in\bigcap_{f\in I}\mathcal{Z}[f]$ and so $I\subseteq \frak{m}\!$$_x$.
So, if $I$ is maximal then we have that $I=\frak{m}\!$$_x$.

$\textbf{Proof (2):}$ If we define $\frak{o}\!$$_x:=\{f\in\mathcal{C}|f \text{ vanishes on a neighborhood of }x\}$, then this is an ideal of $\mathcal{C}$. Clearly $\frak{o}\!$$_x\subsetneq\frak{m}\!$$_x$, but if we take $y\neq x$ there must be a neighborhood $U$ of $x$ such that $y\notin\overline U$ because $M$ is Hausdorff. Because compact and Hausdorff implies Tikhonov, we know that there must be $f\in\mathcal{C}$ such that $U\subseteq\mathcal{Z}[f]$ and $f(y)=1$. Then we may conclude that $\frak{o}\!$$_x\not\subseteq\frak{m}\!$$_y$.