If we assume the Axiom of Choice (equivalently, every proper ideal in a ring with unity is contained in a maximal ideal), then the following works:
Let $F$ be any field, and let $E$ be an infinite set. For each $f\in F^E$, define the support of $f$ to be
$$\mathrm{supp}(f) = \{e\in E\mid f(e)\neq 0\}.$$
Let $I=\{f\in F^E\mid |\mathrm{supp}(f)|\text{ is finite}\}$ be the ideal of functions of finite support.
Note that this is indeed an ideal, since $\mathrm{supp}(f+g)\subseteq\mathrm{supp}(f)\cup\mathrm{supp}(g)$, $\mathrm{supp}(fg)= \mathrm{supp}(f)\cap\mathrm{supp}(g)$, and the zero function has empty, hence finite, support.
On the other hand, $I$ is not all of $F^E$, since the constant function $f(e)=1$ is not in $I$, as we are assuming that $E$ is infinite.
By Zorn's Lemma, $I$ is contained in a maximal ideal $M$ of $F^E$. I claim that for all $a\in E$, $M\neq M_a$.
Indeed, let $a\in E$. Then the function $\chi_{\{a\}}$, given by
$$\chi_{\{a\}} (e) = \left\{\begin{array}{ll}
1 &\text{if }e\in\{a\},\\
0 &\text{if }e\notin\{ a\}
\end{array}\right. = \left\{\begin{array}{ll}
1 & \text{if }e=a,\\
0 & \text{if }e\neq a,
\end{array}\right.$$
lies in $I$, and hence in $M$. But $\chi_{\{a\}}\notin M_a$. Thus, $M\neq M_a$. This holds for all $a\in E$, so $M$ is a maximal ideal that is different from all ideals of the form $M_a$.
Can we do this without the Axiom of Choice? No; we need at least some choice. Recall that given a set $X$, a filter on $X$ is a nonempty collection $\mathscr{F}$ of subsets of $X$ such that (i) if $A\in\mathscr{F}$ and $A\subseteq B$, then $B\in\mathscr{F}$; and (ii) If $A_1,A_2\in\mathscr{F}$, then $A_1\cap A_2\in\mathscr{F}$. An ultrafilter on $X$ is a filter with the property that for all subsets $A$ of $X$, either $A\in\mathscr{F}$ or $X\setminus A\in\mathscr{F}$ (equivalently, $\mathscr{F}$ is maximal among proper filters, partially ordered by inclusion, where the improper filter is the collection of all subsets of $X$).
Lemma. Let $\mathscr{G}$ be a filter on $E$. Then
$$I(\mathscr{G}) = \{f\in F^E\mid E\setminus\mathrm{supp}(f)\in\mathscr{G}\}$$
is an ideal of $F^E$, and is a proper ideal if and only if $\varnothing\notin\mathscr{G}$.
Proof. The zero function lies in $I(\mathscr{G})$. If $f,g\in I(\mathscr{G})$, then $\mathrm{supp}(f+g)\subseteq \mathrm{supp}(f)\cup\mathrm{supp}(g)$, so
$$E\setminus\mathrm{supp}(f+g)\supseteq (E\setminus\mathrm{supp}(f))\cap (E\setminus\mathrm{supp}(g)),$$
hence it contains the intersection of two elements of $I(\mathscr{G})$ and so lies in $\mathscr{G}$. Finally, if $f\in I(\mathscr{G})$, and $g\in F^E$, then $\mathrm{supp}(gf)\subseteq \mathrm{supp}(f)$, so $gf\in I(\mathscr{G})$.
The ideal $I(\mathscr{G})$ is proper if and only if it does not contain any units, if and only if it does not contain any element whose support is all of $E$, if and only if the empty set is not in $\mathscr{G}$. $\Box$
Lemma. Let $I$ be an ideal of $F^E$. Then the collection
$$\mathscr{F}(I)=\{E\setminus \mathrm{supp}(f)\mid f\in I\}$$
is a filter on $E$. If $M$ is maximal, then $\mathscr{F}(M)$ is an ultrafilter on $E$.
Proof. Note that $I$ is nonempty, so $\mathscr{F}(I)$ is nonempty.
Let $A\in\mathscr{F}(I)$, and let $A\subseteq B\subseteq E$. Then there exists $f\in I$ such that $\mathrm{supp}(f)=E\setminus A$. Let $g=\chi_{E\setminus B}$ be the characteristic function of $E\setminus B$. Then $\mathrm{supp}(g)=E\setminus B$. So
$$\begin{align*}
E\setminus \mathrm{supp}(fg) &= E\setminus(\mathrm{supp}(f)\cap\mathrm{supp}(g))\\
&= (E\setminus \mathrm{supp}(f))\cup(E\setminus\mathrm{supp}(g))\\
&= A\cup (E\setminus(E\setminus B))\\
&= A\cup B = B.
\end{align*}$$
Since $fg\in I$, then $B\in\mathscr{F}(I)$.
Finally, if $A_1,A_2\in \mathscr{F}(I)$, then there exist functions $f_1,f_2\in I$ such that $\mathrm{supp}(f_1)=E\setminus A_1$ and $\mathrm{supp}(f_2)=E\setminus A_2$. Let $g=\chi_{A_1} f_2$, which lies in $I$. Note that $\mathrm{supp}(gf_2) = A_1\cap(E\setminus A_2) = A_1\setminus A_2$. Thus, the support of $gf_2$ and the support of $f_1$ are disjoint, so
$$\mathrm{supp}(f_1+gf_2) = \mathrm{supp}(f_1)\cup\mathrm{supp}(gf_2) = (E\setminus A_1)\cup (A_1\setminus A_2).$$
Therefore
$$\begin{align*}
E\setminus\mathrm{supp}(f_1+gf_2) &= E\setminus ( (E\setminus A_1)\cup (A_1\setminus A_2))\\
&= A_1\cap E\setminus(A_1\cap (E\setminus A_2))\\
&= A_1\cap ((E\setminus A_1)\cup A_2)\\
&= (A_1\cap (E\setminus A_1)) \cup (A_1\cap A_2)\\
&= \varnothing\cup (A_1\cap A_2)\\
&= A_1\cap A_2.
\end{align*}$$
Thus, $A_1\cap A_2\in \mathscr{F}(I)$. Therefore, $\mathscr{F}(I)$ is a filter.
Now assume that $\mathscr{F}(I)$ is not an ultrafilter on $E$. Let $A\subseteq E$ be such that $A\notin\mathscr{F}(I)$ and $E\setminus A\notin\mathscr{F}(I)$. We know we can extend $\mathscr{F}(I)$ to a filter $\mathscr{G}$ such that $\mathscr{F}(I),A\subseteq \mathscr{G}$, namely consider the collection of all subsets of $E$ that contain the intersection of an element of $\mathscr{F}(I)$ with $A$. This will not include the empty set because we are assuming that $E\setminus A$ is not in $\mathscr{F}(I)$, and is easy to verify that this is a filter on $E$.
Now consider $I(\mathscr{G})$. Then because $\mathscr{F}(I)\subseteq \mathscr{G}$, if $f\in I$, then $f\in I(\mathscr{G})$; and the characteristic function of $X\setminus A$ lies in $I(\mathscr{G})$, but not in $F$ (since $X\setminus A\notin \mathscr{F}$). Thus, $I$ is not a maximal ideal in $F^E$, because it is properly contained in $I(\mathscr{G})$.
Thus we've shown that if $\mathscr{F}(I)$ is not an ultrafilter, then $I$ is not maximal. Thus, if $M$ is maximal, then $\mathscr{F}(I)$ is an ultrafilter on $E$. $\Box$
An ultrafilter $\mathscr{F}$ on $E$ is principal if there exists $a\in E$ such that $A\in\mathscr{F}\iff a\in A$ (these are easily seen to be ultrafilters). So the existence of a maximal ideal of $F^E$ that is not of the form $M_a$ for some $a\in E$ is equivalent to the existence of a nonprincipal ultrafilter on $E$.
However, one cannot prove the existence of a nonprincipal ultrafilter on a (necessarily infinite) set without some choice. We can't even do it in a nice set like $\omega$ with its well ordering. So you won't be able to come up with any example that is explicitly cnostructed, as that would allow you to explicitly construct a nonprincipal ultrafilter on an infinite set.
