It looks like you must intend this for commutative rings, because it doesn't hold for general noncommutative rings.
The answer to the question follows from the same argument here in Manny Reyes's solution.
If the faithful and finite length module is $M$, then the trick is just to embed $R$ into $M^n$. As a submodule of an Artinian module, it is also then Artinian.
Added more detail: Since $M$ has finite length, it has a finite generating set $m_1,\dots m_n$. Look at the map $R\to M^n$ given by $r\mapsto (m_1r,m_2r,m_3r,\dots,m_nr)$. Since $M$ is faithful, this map is an injection. Since $M$ is Artinian, so is $M^n$. Thus, $R$ is a submodule of an Artinian module, and hence is Artinian itself.
The converse is obvious since if $R$ is Artinian, then it itself is a faithful $R$ module of finite length.