A friend and I are having a disagreement.
I think that $(x+1)(x-1) = 0 \in \mathbb{Z}_3[X] /(X^2 + 2)$ Which means that there are zero divisors in $\mathbb{Z}_3[X] /(X^2 + 2)$. There is no equivalent in $\mathbb{Z}_3 \times \mathbb{Z}_3$. Therefore they cannot be isomorphic.
However, I may be misunderstanding the elements of $\mathbb{Z}_3[X] /(X^2 + 2)$.
What are the elements of $\mathbb{Z}_3[X] /(X^2 + 2)$? Are there 9? Are there zero divisors?
The two rings are indeed isomorphic.
As pointed out in the comments, $(X^2 -1) = (X+1)(X-1)$. We also have that the ideals $I = (X+1)$ and $J = (X-1)$ are comaximal. Indeed, since $(X+1)-(X-1) = 2$ is a unit, we have that $I + J = \mathbb{Z}_3[X]$. We can now conclude by the Chinese Remainder Theorem: $$\mathbb{Z}_3[X]/(X^2 -1) \cong \mathbb{Z}_3[X]/(X+1) \times \mathbb{Z}_3[X]/(X-1) \cong \mathbb{Z}_3 \times \mathbb{Z}_3$$
Explicitly, we can write the isomorphism $\mathbb{Z}_3[X]/(X^2 -1) \to \mathbb{Z}_3 \times \mathbb{Z}_3$ as $X \mapsto (-1,1)$.