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Let $R$ be a commutative unital ring with just one maximal ideal $I$. Then $R\backslash I \subset R^\times$, where $R^\times$ is the set of units of $R$.

My approach:

Since $I$ is maximal and $R$ is commutative, then $I$ is prime. Since $I$ is a subgroup of $R$, and $R$ is unital, $1\in I$, so $1\not\in R\backslash I$. Now, if $r\in R\backslash I$ then $r=1r\in I$ (since $I$ is prime), a contradiction. Hence $R\backslash I$ is empty (since $0$ is also in $I$). Therefore, $R\backslash I\subset R^\times$.

Do you think this is correct, or am I mistaken somewhere?

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    $I$ is an *additive* subgroup of $R$, so it's not true that $1\in I$. Generally an ideal is not called maximal unless $I\subsetneq R$. – 2017-02-13
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    **Hint** Let $a \in R \setminus I$. Consider the ideal $(a)$. Can it be a *proper* ideal? If it is, then it is contained in... – 2017-02-13
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    @AndreasCaranti I think it can't be a proper ideal, since if we're just multiplying all elements in $R$ by $a$ and we always get some element in $R$, and $a\ne 0$. Unless $\exists r_1\ne r_2$ such that $ar_1 = ar_2$, which would mean that $R$ is not an integral domain. – 2017-02-13
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    The reason it can't be a proper ideal is that if ot were, then it would be contained in $I$. But $a \in (a)$, while $a \notin I$. – 2017-02-13

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Let $a \in R \setminus I $, then the ideal $(a)$ is not contained in $I$, as $I$ is the only maximal ideal, this means that $(a)$ is not contained in any maximal ideal. As any proper ideal is contained in a maximal ideal, we must have $(a) = R$. Thus $a \ \in R^\times$.

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    You should indicate that "any proper ideal is contained in a maximal ideal" is [Krull's theorem](https://en.wikipedia.org/wiki/Krull's_theorem). – 2017-02-13