In this paper Mel Henriksen shows that a commutative ring $R$ has no maximal ideals iff (a) $J(R)=R$, where $J(R)$ is the Jacobson radical of $R$, and (b) $R^2+pR=R$ for every prime $p\in\Bbb Z$. He then gives three examples. One starts with a field $F$ of characteristic $0$ and forms the integral domain
$$S(F)=\left\{h(x)=\frac{f(x)}{g(x)}\in F(x):f(x),g(x)\in F[x]\text{ and }g(x)\ne 0\right\}\;;$$ its unique maximal ideal is $R(F)=xS(F)$, which has no maximal ideals.
This paper by Patrick J. Morandi also constructs some examples.
What you say is correct and is true in more generality. You should be interested by the following result:
Let $K$ be a topological compact space. Then, the maximal ideals of $A
:= \mathcal{C}(K, \mathbb{R})$ are exactly the ideals of functions
vanishing at a fixed point of $K$.
Here are some hints to prove this theorem:
1) Determine the invertibles of $A$ (easy).
2) For any $x$ in $K$, show that the set of functions of $A$ vanishing in $x$ is a maximal ideal of $A$ (easy).
3) Conversely, let $I$ be a maximal ideal of $A$. Suppose by contradiction that for any $x$ in $K$, you can find a function $f_x$ in $A$ which does not vanish at $x$. By continuity, there is a neighborhood $U_x$ of $x$ such that $f_x$ does not vanish on $U_x$.
4) Using compactness, construct a function of $A$ that vanishes nowhere and conclude (clever). Hint: in the real numbers, a sum $\sum \lambda_i^2$ vanishes if, and only if, all $\lambda_i$ vanish.
Remark: adapt the proof, for functions with values in $\mathbb{C}$.
Edit: concerning the question of primality of ideals of functions vanishing at more than one point, you can prove this as follows:
Let $I_C$ be the set of functions of $A$ vanishing one some closed set $C$ of $[0,1]$. Write $C = X \cup Y$ where $X$ and $Y$ are closed proper subsets of $[0,1]$ in $C$ (this is always possible if $C$ has at least two points x < y: take for $X$ the set of points $t \in A$ with $t \leq \frac{x+y}{2}$ and for $Y$ the set of points $t \in A$ with $t \geq \frac{x+y}{2}$).
Let $f$ and $g$ be functions in $A$ vanishing exactly on $X$ and $Y$. For instance, you can take the distance functions to $X$ and $Y$. Then $f.g$ is in $I_C$ but neither $f$ nor $g$ are in $I_C$.
This can certainly be generalized for a more general topological space, but maybe the compactness condition is not sufficient.
Best Answer
No.
Let $f(z)=1/\Gamma(z)$ where $\Gamma$ is the gamma function. Then $f$ vanishes at $0$, $-1,-2,\ldots$ and nowhere else. For integers $m$ let $f_m(z)=f(z+m)$. Then $f_m$ vanishes at $-m,-m-1,\ldots$. Let $I$ be the ideal of the ring of the entire functions generated by the $f_m$. Then $I$ is proper, since otherwise finitely many of the $f_m$ would generate an ideal containing the constant function $1$. But any finite set of $f_m$ have a common zero, so this doesn't hold. So $I\subseteq J$ for some maximal ideal $J$ (using the standard Zorn's lemma argument). But $J\ne M_\lambda$ for any $\lambda$, as for each $\lambda$ there is some $m$ with $f_m(\lambda)\ne0$, and $f_m\in J$.