Question: Let $R$ be a nonzero commutative ring with $1$. If $I$ is an ideal of $R$ such that $1+a$ is a unit in $R$ for all $a\in I$ then $I$ is contained in every maximal ideal of $R$.
My apporoach: This question is asked here Show that $I$ is contained in every maximal ideal of $R$, but I approach it a similar, but different way, and I just want to see if it works:
Suppose $M$, maximal, which does not contain $a$. Thus, $M\subset I+M=R$. Let $b\in M$, then $a+b=1$, since $1\in R$. Then, $b=1-a\in 1+(a)\in M$, a contradiction. Hence, $I\subseteq M$, as wanted.
I suppose I couldn't conclude the problem based on the hint, but I'm sure there was only one or two more lines.
Key: If an ideal of $R$ contains a unit, the ideal must be $R$.
Proof: Suppose that there exists a maximal ideal $\mathfrak m$ which does not contain $I$. Then $\mathfrak m$ is properly contained in $I+\mathfrak m$, and then $I+\mathfrak m = R$ by maximality. Thus we can find $a\in I$ and $b \in \mathfrak m$ such that $1=a+b$. By assumption, $b=1-a$ is a unit, so $\mathfrak m = R$, and this is a contradiction.