This is a CW answer composed in an effort to remove this question from the unanswered queue. It is based on Bruno's suggestions in the comments.
Suppose $R$ is a semisimple Artinian ring, which can be expressed as a finite direct sum $R\cong \oplus R_i$ where the $R_i$ are simple Artinian rings. For any right $R$ module $M$, $MR_i$ is a right $R$ submodule of $M$ (as well as a right $R_i$ module), and moreover $M=\oplus MR_i$. This is just the decomposition of $M$ into homogeneous pieces, and indeed the distinct $MR_i$ are sums of distinct isotypes of simple $R$-submodules of $M$.
The decomposition also establishes that every submodule of $M$ is going to be a direct sum of submodules of the $MR_i$. So, the product of these $R$ modules will have finitely many submodules if each individual $MR_i$ has finitely many submodules.
We will show that for a simple Artinian $\Bbb C$-algebra $A$, an $A$-module has finitely many submodules iff the module is simple. Consider $S\oplus S$ for a simple $A$-module $S$. Observe that $(x,\lambda x)A$ is a submodule of $S\oplus S$ for any $\lambda\in \Bbb C$ and $x\neq 0$.
Now if $(x,\alpha a)A=(x,\beta x)A$ for some $\alpha,\beta$, it follows that $(xY,\alpha xY)=(x,\beta x)$ for some $Y\in A$. But then $x= xY$, and $\alpha x=\alpha x Y=\beta x$. Since $x\neq 0$, $\alpha=\beta$. This shows that for distinct choices of $\lambda$, $(x,\lambda x)A$ are distinct submodules. Since there are infinitely many choices for $\lambda$, $S\oplus S$ has infinitely many submodules. Any nonsimple $A$ module would have to contain at least two copies of $S$, and hence infinintely many submodules.
So in summary, if $M$ has finitely many submodules, then each of the $MR_i$ is actually a simple $R$ (and simple $R_i$) module. Since each $MR_i$ corresponds to a distinct homogeneous piece of $R$, they are pairwise nonisomorphic. In this case it is even possible to count the submodules: if there are, say, $k$ of the $MR_i$ which are nonzero in the decomposition, then $M$ has $2^k$ submodules