Can the tensor product of two commutative rings of characteristic zero (over the integers $\mathbb{Z}$) be trivial, or at least have nonzero characteristic?
Note that any nontrivial ring with a torsion-free additive group has characteristic zero.
If $R$ is a nontrivial ring with torsion-free additive group and $S$ has characteristic zero, then $R \cong R \otimes \mathbb{Z} \to R \otimes S$ is injective because $R$ is a flat $\mathbb{Z}$-module and $\mathbb{Z} \to S$ is injective, so $R \otimes S$ has characteristic zero and is a fortiori nontrivial.
Hence, any answer to the question must involve rings that have non-torsion-free additive groups, and cannot involve integral domains.
Let $T(R)$ denote the ideal of torsion elements of $R$. Then $R/T(R)$ is torsion-free, and nonzero if $R$ has characteristic zero. In particular, then, if $R$ and $S$ have characteristic zero, $R/T(R)\otimes S/T(S)$ is a ring of characteristic zero by your flatness argument. Since there is a natural surjection $R\otimes S\to R/T(R)\otimes S/T(S)$, this implies $R\otimes S$ also has characteristic zero.
Alternatively, you can use the fact that $R$ has characteristic zero iff $R\otimes\mathbb{Q}$ is nonzero. If $R$ and $S$ have characteristic zero then $(R\otimes S)\otimes \mathbb{Q}\cong(R\otimes\mathbb{Q})\otimes_\mathbb{Q} (S\otimes\mathbb{Q})$ is nonzero since it is a tensor product of two nonzero $\mathbb{Q}$-vector spaces, so $R\otimes S$ has characteristic zero.