= Solution
In a Boolean <ring>, $(r+1)^2=r+1$. Using $r^2=r$ and $1^2=1$ gives
$$
r+2r+1=r+1,
$$
so
$$
\boxed{2r=0\quad\text{for every }r\in R}.
$$
If $R$ is also a nonzero <integral> domain, then
$$
r(r-1)=r^2-r=0
$$
forces every $r$ to equal $0$ or $1$. Hence $R\cong\mathbb F_2$.
If $I$ is prime in a Boolean <ring>, then $R/I$ is a nonzero Boolean <integral> domain by part (i), and hence is $\mathbb F_2$. Since the quotient is a field,
$$
\boxed{I\text{ is maximal}}.
$$
This is the <prime ideals of a Boolean ring are maximal> property.
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