I've been thinking about this too, see here, so here are a few examples to help you understand the definitions better:
$\mathbb Z$:
Note that $\mathbb Z$ is a $\mathbb Z$-module. We claim that it is Noetherian. To see this, note that submodules of $\mathbb Z$ correspond to subgroups of it so that all submodules are of the form $(n)$, the ideal generated by $n$ since $\mathbb Z$ is a PID. Now if you have $(n) \subset (m)$ then $m$ must divide $n$ so that $(n_1) \subset (n_2) \subset (n_3) \subset \dots$ stabilises for $k$ large enough since $n_1$ only has a finite number of divisors. Hence every ascending chain in $\mathbb Z$ must be stationary which means that $\mathbb Z$ is Noetherian.
On the other hand, observe that for any $n > 1$ you have $(n) \supset (n^2) \supset (n^3) \supset \dots$ which is a decreasing chain of submodules and hence $\mathbb Z$ is not Artinian.
A finite abelian group $G$:
Since $G$ only has finitely many elements, every increasing and every decreasing sequence will eventually be stationary and hence a finite abelian group is always both, Artinian and Noetherian.
$\mathbb Q$ (as a $\mathbb Z$-module):
Note that $\mathbb Q$ contains $\mathbb Z$ as a submodule hence $\mathbb Q$ cannot be Artinian.
Also note that if $(\frac1p)$ denotes the submodule of $\mathbb Q$ generated by $\frac1p$ then the following is an increasing sequence of submodules (and hence $\mathbb Q$ is not Noetherian): $(\frac1p) \subset (\frac{1}{p^2}) \subset (\frac{1}{p^3}) \subset \dots$
$\mathbb Q / \mathbb Z$ (as $\mathbb Z$-module):
This is not Noetherian since for example $(\frac1p) \subset (\frac{1}{p^2}) \subset \dots$ is an increasing chain of submodules that is not stationary.
On the other hand, subgroups of $\mathbb Q / \mathbb Z$ look like $(\frac{1}{n})$, the subgroup generated by $\frac1n$. If we have $(\frac1n) \supset (\frac1m)$ we know that $m$ divides $n$ so if $(\frac{1}{n_1}) \supset (\frac{1}{n_2}) \supset \dots$ is a decreasing chain it eventually becomes stationary because there are only a finite number of divisors of $n_1$. Hence $\mathbb Q / \mathbb Z$ is Artinian.
We will use the following standard commutative algebra fact:
Let $A$ be a commutative ring, and $E$ be an $A$-module. Suppose there is a filtration $$E = E_0\supseteq E_1\supseteq\cdots\supseteq E_n = 0$$ of $A$-submodules $E_i$. Then $E$ is Noetherian (resp. Artinian) $\iff$ each quotient $E_i/E_{i+1}$ is Noetherian (resp. Artinian).
Note that you have a filtration $$M \supseteq mM\supseteq m^2M\supseteq\cdots\supseteq m^nM = 0.$$ Thus $M$ is Noetherian (resp. Artinian) if and only if each of the successive quotients $m^kM/m^{k+1}M$ are Noetherian (resp. Artinian) $R$-modules. Note that each quotient $m^kM/m^{k+1}M$ is naturally an $R/m$-module, and moreover that $m^kM/m^{k+1}M$ will be a Noetherian (resp Artinian) $R$-module $\iff$ it is a Noetherian (resp Artinian) $R/m$-module. Since $R/m$ is a field, we know that $m^kM/m^{k+1}M$ is a Noetherian $R/m$-module $\iff$ it is an Artinian $R/m$-module $\iff$ it is finite dimensional over $R/m$. In particular, each of the quotients $m^kM/m^{k+1}M$ is Noetherian if and only if it is Artinian, and hence $M$ is Noetherian if and only if it is Artinian.
Best Answer
Let $A$ be an integral domain of dimension greater than 1. Then $A_{(0)}$ is a field, so in particular it is artinian. But $A$ is not artinian.