What are the rings of order $8$ that don't have nonzero nilpotent elements? I am sure those are $3$ rings. Are those $\Bbb{F}_2\times \Bbb{F}_4$, $\Bbb{F}_8$ and $\Bbb{F}_2\times\Bbb{F}_2\times\Bbb{F}_2$?
 A: Yes, that's all. (I'm assuming the rings have identity, too.)
One way to see that this list is complete is to say that finite, reduced rings are semisimple, and hence they are products of matrix rings over division rings (fields, actually, since the ring is finite.)
Since the ring is commutative, the matrix rings all have to be trivial size ($1\times 1$ matrices), so we are looking at a product of fields.
The only possible sizes of fields have to divide $8$, and you have found all the combinations that are possible to do that.
A: Due to the tag "commutative algebra", I will assume that "ring" means commutative ring.
There is nothing special about the number $8$ here, so I will classify reduced commutative rings of finite order.
We can prove the following lemma in more generality:

Lemma If $R$ is a commutative Artinian reduced ring, then $R$ is isomorphic to a finite product of fields.

Proof Since $R$ is Artinian, $R$ has finitely many prime ideals and all prime ideals are maximal. Let $\mathfrak{p}_1, \dots, \mathfrak{p}_k$ all the prime ideals. We have $\displaystyle \bigcap_{i=1}^k \mathfrak{p}_i=\sqrt{(0)}=(0)$ and $\mathfrak{p}_i + \mathfrak{p}_j = R$ for $i \neq j$ (due to the maximality of $\mathfrak{p}_i$ and $\mathfrak{p}_j$), thus the Chinese remainder theorem gives an isomorphism $$R \cong R/(0) \cong R/(\cap_{i=1}^k \mathfrak{p}_i) \cong \prod_{i=1}^k R/\mathfrak{p}_i$$
Where each $R/\mathfrak{p}_i$ is a field, which proves the claim.
Of course, every finite ring is Artinian, so we obtain

Corollary Every (commutative) reduced finite ring is a finite product of finite fields.

Together with the classification of finite fields (i.e. there is exactly one for every prime power), one can figure out the possible reduced commutative rings of a certain order $n$ by some elementary combinatorics. 
