= Solution
{c}
\b[No.] A <domain with a nonzero idempotent maximal ideal> is a counterexample. Consider
$$
R=\bigcup_{n\geq0}\mathbb Q[t^{1/2^n}],\qquad
u_n=t^{1/2^n},\qquad u_n=u_{n+1}^2.
$$
This can be realized concretely as a <ring of functions> on $t>0$: its elements are finite sums of rational multiples of nonnegative <dyadic rational> powers of $t$. Each stage $\mathbb Q[u_n]$ is a <polynomial ring>, because a nonzero <polynomial> cannot vanish for all $u_n>0$. Any two elements lie in one stage, which is an <integral domain>, so their product is nonzero if both are nonzero. Thus $R$ is an <integral domain>.
The constant term defines a <surjective> <ring homomorphism> $\epsilon:R\to\mathbb Q$. It is well defined because replacing $u_n$ by $u_{n+1}^2$ preserves the constant term. Its <kernel of a ring homomorphism>
$$
\mathfrak m=\{f\in R:\text{the constant term of }f\text{ is }0\}
$$
is a <maximal ideal>, since $R/\mathfrak m\cong\mathbb Q$ is a <field>, by the <maximal ideal quotient criterion>. Also $u_0=t\in\mathfrak m$ is nonzero.
If $f\in\mathfrak m$, choose a stage in which $f=u_ng(u_n)$ for a <polynomial> $g\in\mathbb Q[z]$. Then
$$
f=u_{n+1}\bigl(u_{n+1}g(u_{n+1}^2)\bigr).
$$
Both factors belong to $\mathfrak m$, so $f\in\mathfrak m^2$. The reverse inclusion holds for every <ideal>, proving that $\mathfrak m$ is an <idempotent ideal>. Hence
$$
\boxed{0\ne\mathfrak m\text{ is maximal},\qquad\mathfrak m^n=\mathfrak m^{n+1}=\mathfrak m\quad(n\geq1).}
$$
The <Noetherian ring> hypothesis in <2/c> is therefore essential even for a <maximal ideal>.
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