Suppose ${\rm rank}\, x = \alpha$, ${\rm rank}\,y = \beta$ and $\alpha \le \beta$. Then ${\rm rank}\,(x,y) = \beta + 2$.
I can prove this for $\alpha=\beta$, and I have a proof for $\alpha \le \beta$, but I'm not sure that it's correct. Here goes:
By definition, $\alpha$ is the smallest ordinal for which $x \in V_\alpha$ and $\beta$ is the smallest ordinal with $y \in V_\beta$. We wish to form the set $(x,y) =\{ \{x\} , \{x, y\} \}$ and to determine the smallest $\gamma$ for which $(x,y) \in V_\gamma$. Since the rank cannot be a limit ordinal, there is only the case in which $\beta$ is a successor ordinal.
Since $\alpha \le \beta$, then $V_\alpha \subseteq V_\beta$ (set equality when $\alpha=\beta$), and therefore $x \in V_\beta$. This means $\{x, y\} \subseteq V_\beta$, and so $\{x, y\} \in V_{\beta + 1}$. In a similar manner, $\{x\} \subseteq V_\alpha \subseteq V_\beta$, which means $\{x\} \in V_{\beta+1}$. Thus $(x,y) \subseteq V_{\beta+1}$, and so $(x,y) \in V_{\beta+2}$. While $\{x\} \in V_{\alpha + 1}$, the pair $\{x,y\} \in V_{\beta+1}$ since ${\rm rank}\,y = \beta$. Therefore $V_{\beta+2}$ is the earliest set in the hierarchy that contains $(x,y)$.
Have I made a mistake? The proof looks right, but it feels wrong.