 Solution

If the quotient ring $R/I$ is an integral domain, then $I\ne R$, because an integral domain has $1\ne 0$. For $ab\in I$,There are no nonzero zero divisors in an integral domain, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving $I$ is a prime ideal.

If $R/I$ is a field, again $I$ is proper. Let $J$ be an ideal properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a multiplicative inverse $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore $I$ is a maximal ideal. These arguments establish the relevant prime ideal quotient criterion and maximal ideal quotient criterion.