 Solution

Let $L$ be any maximal left ideal. If $I⊈L$, maximality gives $L+I=R$, because the two-sided ideal $I$ is in particular a left ideal. Thus $1=\ell +i$ for some $\ell \in L$ and $i\in I$. Butis a unit by hypothesis. A proper left ideal cannot contain a unit, since left multiplication by its inverse would put $1$ in the left ideal. This contradiction proves $I\subseteq L$ for every maximal left ideal $L$, and thereforeTogether with ii, this is the ideal form of the unit criterion for the Jacobson radical.