**Proof plan.** Use the correspondence between ideals of $R/I$ and ideals of $R$ containing $I$. A field has only two ideals; a ring with only two ideals is a field. **Step 1: ($\Rightarrow$) If $I$ is maximal, then $R/I$ is a field.** [claim: Maximal Implies Field] If $I$ is maximal, then $R/I$ has no proper non-zero ideals, hence is a field. [/claim] [proof] The ideal correspondence theorem states that ideals of $R/I$ correspond bijectively to ideals of $R$ containing $I$. If $I$ is maximal, the only ideals of $R$ containing $I$ are $I$ itself and $R$. These correspond to $\{0\}$ and $R/I$ in $R/I$. So $R/I$ has only the two trivial ideals. Since $R/I \neq 0$ (as $I \neq R$), every non-zero element generates the whole ring, so it is a unit. Hence $R/I$ is a field. [/proof] **Step 2: ($\Leftarrow$) If $R/I$ is a field, then $I$ is maximal.** [claim: Field Implies Maximal] If $R/I$ is a field, then $I$ is maximal. [/claim] [proof] A field has only two ideals: $\{0\}$ and itself. By the ideal correspondence theorem, the only ideals of $R$ containing $I$ are $I$ and $R$. This is precisely the definition of maximality. [/proof] These two claims together give the equivalence. $\square$