In R. Ash, Abstract Algebra in the chapter on non-commutative rings (9.2.2), the following exericse occurs:
Let $M$ be a nonzero cyclic module. Show that $M$ is simple if and only if $\operatorname{ann} M$, the annihilator of $M$, is a maximal left ideal.
As the chapter says, the rings are not assumed to be commutative, the modules are assumed to be left modules, and the annihilator is the set of all $r \in R$ such that $rm = 0$ for all $m \in M$.
Now I tried messing around with the map $f_x(r) = rx$ for $x \ne 0$, which is surjective by simplicity. Then I tried looking at its kernel, and surely we have $\operatorname{ann} M \subseteq \operatorname{ker}(f)$. But I do not see the converse inclusion (which should hold according to the solution given, which I looked up). If $R$ is not commutative, we cannot conclude from $rx = 0$ that $r(sx) = 0$ for some $s \in R$? All I see is that $$ \operatorname{ann}(M) = \bigcap_{x\in M} \operatorname{ker}(f_x) $$ and that if $y = sx$ we have $r \in \mbox{ker}(f_y) \Leftrightarrow ry = 0 \Leftrightarrow (rs)x = 0 \Leftrightarrow rs \in \mbox{ker}(f_x)$ hence $\mbox{ker}(f_{sx}) = \{ r \in R \mid rs \in \mbox{ker}(f_x) \}$. But I do not see that the annihilator coincides with any kernel?
Let $R=M_2(\mathbb R)$ and $L$ be any nonzero left submodule. Then the left annihilator of $L$ is $\{0\}$ since $R$ is a simple ring, and in particular the annihilator is not a maximal left ideal.
This is true even for the $L$'s which are simple, such as $\left\{\begin{bmatrix}x&0\\y&0\end{bmatrix}\middle|\, x,y\in\mathbb R\right\}$.
So the statement is false.
But if $S$ is any simple left $R$ module, $S$ is cyclic, so you can write $S=Rx$ for some $x$ (and in fact, any nonzero element of $S$ will do). Then you have a natural homomorphism $R\to Rx$ given by $r\mapsto rx$, whose kernel is a maximal left ideal of $R$ by the first homomorphism theorem.
So, the left annihilator of any nonzero element of a simple left module is a maximal left ideal.