Solution: Recall that a maximal ideal of $R$ is a proper ideal $I$ of $R$ such that for any ideal $J$ of $R$, if $I\subseteq J$ and $J\neq R$ then $I=J$.

Let $R$ be a field. It is easy to see that $\{0\}$ is an ideal of $R$. We will prove that $\{0\}$ is the only proper ideal, which in particular prove that $\{0\}$ is a maximal ideal. For that, let $I\neq \{0\}$ be an ideal of $R$. Take a non-zero element, say $x\in I$. Since $R$ is a field, $x^{-1}\in R$, so $x x^{-1}=1\in I$. Since $1\in I$, then for any $r\in R$ $1r=r\in I$ and hence $I=R$. Therefore, $\{0\}$ is a maximal ideal of $R$.

On the other hand, let $\{0\}$ be a maximal ideal of $R$. We need to show that $R$ is a field. Recall that, a commutative ring $R$ with unity is a field if non-zero element is invertible. Let $r\in R$ be a non-zero element. Consider the ideal $\left\langle r \right\rangle$. Since $\{0\}$ is a maximal ideal, we must have $\left\langle r \right\rangle =R$. Since $1\in R = \left\langle r \right\rangle$, there exists $s\in R$ such that $rs=1$. Thus, $r$ is invertible and hence every non-zero element is invertible. Hence, $R$ is a field.