Let $R$ be a ring such that every left $R$-module is free, and let $I \subset R$ be a maximal left ideal. Then $R/I$ is a simple nonzero $R$-module, and is free by hypothesis, so $R/I$ has a basis. Take any basis element $x$, and let $\varphi \colon R \to R/I$ be the $R$-module homomorphism given by $\varphi(r) = rx$. Since $x$ is nonzero and $R/I$ is simple, $Rx = R/I$, so $\varphi$ is surjective. On the other hand, $\varphi$ must be injective, as $x$ is a basis element, so $r\cdot x \neq 0$ for any nonzero $r \in R$. Hence, $R \cong R/I$ as $R$-modules, so $R$ must also be simple; in particular, it has no nonzero proper left-ideals.