Eric's result for commutative rings, that an example must have the ring no bigger than the continuum and the maximal ideal infinitely generated, extends to left ideals for non-commutative rings, assuming that by "not a direct summand" you mean "not a direct summand as a left ideal".
Let $A$ be the ring, $I$ the countable maximal left ideal, and $S$ the simple left module $A/I$. Considering these all as left $A$-modules, there can be no non-zero maps $S\to I$ or $I\to S$. It follows that there can be no non-zero map $S\to A$, since if there were, then composing with the natural epimorphism $A\to S$ would give a non-zero map $S\to S$, which must be invertible since $S$ is simple, and so $A\cong I\oplus S$ as left $A$-modules.
Hence the map $A\to A$ given by right multiplication by any $0\neq a\in A$ must restrict to a non-zero map $I\to A$, and therefore to a non-zero map $I\to I$, since there are no non-zero maps $I\to S$.
So, as in Eric's argument, $A$ acts faithfully on $I$.
EDIT: By the way, this shows that $I$ is a two-sided ideal. The quotient ring $S=A/I$ must then be a skew field, since it is simple as a left module for itself. Also, $I/I^2$ is an $S$-module, but is at most countable, so must be zero. So $I$ is an idempotent ideal.