$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.

As you noted let $\frac{A}{H} \trianglelefteq \frac{G}{H}$ wherein $H \trianglelefteq A\trianglelefteq G.$ Since $H$ is a maximal subgroup, $H=A$ or $A=G$ and so, $\frac{A}{H}=1$ or $\frac{A}{H}=\frac{G}{H} $.

This means that $\frac{G}{H} $ is a simple group. Now suppose that $H \trianglelefteq G$ and $\frac{G}{H} $ is simple. If we have $H \trianglelefteq A\trianglelefteq G$ then obviously $\frac{A}{H} \trianglelefteq\frac{G}{H}$ and that $\frac{G}{H}$ is simple, we get $\frac{A}{H}=\frac{G}{H}$ or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$.

I flesh out the exquisite answer of B. S.

$\color{darkred}{ \text{ (1.) If I'm not confounded, I think $H ⊴ G$ means $H ⊲ G$ or $H = G$. Is this perfect? } }$

Forward step: $H \text{ maximal } ⊲ G \implies G/H$ simple.

Let $\frac{A}{H} ⊲\frac{G}{H}$ wherein $H ⊴A⊴G$. Since H is a maximal subgroup, $\begin{cases} H = A \implies \frac{A}{H}=1 \\ \text{ or } A=G \implies \frac{A}{H}=\frac{G}{H} \end{cases}$. This means that $\frac{G}{H} $ is a simple group. ♥

Backward step: Now suppose that $H ⊲G$ and $\frac{G}{H} $ is simple.
If we have $H ⊴A⊴G$ then obviously $\frac{A}{H} ⊲\frac{G}{H}$.
By reason of the presupposition for this backward step, $\frac{G}{H}$ is simple.
Hence $\frac{A}{H}=\frac{G}{H}$ or $\frac{A}{H} =\{H\}$ . So, $A=G$ or $H=A$. ♥