You actually struck upon a good path early on, but then went Off into the weeds.
If you already know that $R/J$ has only finite many maximal ideals, then the next step is to recognize those correspond exactly to the maximal ideals of $R$.
Do you see why? The maximal ideals of $R/I$ are the maximal ideals of $R$ containing $I$. But in the case of $J$...?
The Artin-Wedderburn theorem tells us that the maximal semisimple quotient is a product of matrix rings over finite division rings, one for each irreducible representation. Furthermore, every finite division ring is a field, and the unit group of any finite field is cyclic. The only nontrivial homomorphism from $S_3$ to a cyclic group is the sign homomorphism $S_3\to\mathbb{Z}/2$. It follows that any homomorphism from $\mathbb{Z}S_3$ to a finite field lands in the prime subfield (since elements of $S_3$ can only map to $\pm 1$).
So, writing $\mathbb{F}$ for either $\mathbb{F}_2$ or $\mathbb{F}_3$, the maximal semisimple quotient of $\mathbb{F}S_3$ is a product of matrix rings $M_n(K)$ for finite extensions $K$ of $\mathbb{F}$, one for each irreducible representation, and in all the cases where $n=1$ the $K$ is just $\mathbb{F}$. The only $1$-dimensional representations are the trivial representation and the sign representation, and the sign representation is the same as the trivial representation in the case $\mathbb{F}=\mathbb{F}_2$.
For $\mathbb{F}=\mathbb{F}_3$, dimension-counting now tells us there can be no more irreducible representations: the two $1$-dimensional representations take up $2$ dimensions of the semisimple quotient, and the Jacobson radical is nontrivial since it contains $\sum_{g\in S_3} g$, so there are at most $3$ dimensions left. Another irreducible representation would give a copy of $M_n(\mathbb{F}_{3^d})$ in the semisimple quotient for some $d$ and some $n>1$, which is impossible since there aren't enough dimensions left. We conclude that the two $1$-dimensional representations are the only irreducible representations for $\mathbb{F}=\mathbb{F}_3$, and so the maximal semisimple quotient is $\mathbb{F}_3\times\mathbb{F}_3$. The Jacobson radical is then the kernel of the map $\mathbb{F}_3S_3\to\mathbb{F}_3\times\mathbb{F}_3$; explicitly, it is the set of elements $\sum_{g\in S_3} a_g g$ such that $\sum a_g=0$ and $\sum a_g \sigma(g)=0$, where $\sigma(g)$ is the sign of $g$.
Over $\mathbb{F}_2$, on the other hand, there are up to $4$ dimensions left after accounting for the single $1$-dimensional representation and the fact that the Jacobson radical is nontrivial, so there might be a $2$-dimensional irreducible representation. To find one, note that there is a permutation representation of $S_3$ on $\mathbb{F}_2^3$, and this splits as a direct sum of a trivial subrepresentation (generated by $(1,1,1)$) and a $2$-dimensional subrepresentation (consisting of $(a,b,c)$ such that $a+b+c=0$). (Note that this splitting of the permutation representation doesn't happen over $\mathbb{F}_3$, since $(1,1,1)$ is contained in the latter $2$-dimensional subrepresentation.) This $2$-dimensional representation can easily be verified to be irreducible (for another way of seeing it, note that $\mathbb{F}_2^2\setminus\{0\}$ has three elements, and every permutation of them gives a linear map, so in fact $GL_2(\mathbb{F}_2)\cong S_3$).
So over $\mathbb{F}_2$, we conclude that there is the trivial representation and also this $2$-dimensional irreducible representation; counting dimensions, we now see that we have accounted for all $6$ dimensions of $\mathbb{F}_2S_3$. We conclude that the Jacobson radical is only $1$-dimensional (generated by $\sum_{g\in S_3} g$), and the quotient is $\mathbb{F}_2\times M_2(\mathbb{F}_2)$.
Best Answer
As you said, $R$ is simple, so your maximal ideal $\mathfrak m$ is 0 and it is not prime.