Ok, I did this quickly, it might be wrong:

First, all ideals are of the form $(\pi^i)$. Say $(\pi^n)$ is the annihilator of $M$. We can replace $A$ by $A/(\pi^n)$ wlog. Then $M$ has $A$ as submodule, since there's an element that isn't killed by $(\pi^{n-1})$. Now show that $A$ is injective over itself by Baer's criterion. Say $f : \pi^i A \to A$, we need to extend it to $A$. It suffices to show that $f(\pi^i) \in \pi^i A$. But it follows by induction that if an element of $A$ is $\pi^{n-i}$-torsion, it is a multiple of $\pi^{i}$.

First, all ideals are of the form $(\pi^i)$. Say $(\pi^n)$ is the annihilator of $M$. We can replace $A$ by $A/(\pi^n)$ wlog. Then $M$ has $A$ as submodule, since there's an element that isn't killed by $(\pi^{n-1})$. Now show that $A$ is injective over itself by Baer's criterion. Say $f : \pi^i A \to A$, we need to extend it to $A$. It suffices to show that $f(\pi^i) \in \pi^i A$. But it follows by induction that if an element of $A$ is $\pi^{n-i}$-torsion, it is a multiple of $\pi^{i}$.

Edit: the idea is the same as in Hailong's answer, but you don't need the result of Hungerford.