Extension of Rice's theorem
Let $C$ be the collection of all c.e. sets.
One way to understand Rice's theorem is to say that every computable map $f : C \to \mathsf{bool}$ which preserves equality of sets is constant. (By "respects equality of sets" we mean that $f$ actually accepts codes of c.e. sets, but it maps any two codes for the same set to equal values).

Let's answer your question. Suppose we have a computable map $g : C \times C \to \mathsf{bool}$ which respects equality of sets. We want to prove it is constant. Then for every $y \in C$, the map $x \mapsto g(x,y)$ is constant by Rice's theorem. Likewise, for every $x \in C$, the map $y \mapsto g(x,y)$ is constant by Rice's theorem. But a map $g$ which is constant in each argument separately is constant.