Is a subset of $\mathbb{Q}\times\mathbb{Q}$ that all variants of an exponentiation equation have answers in it, infinite?

Question

Note that we have:

$$A=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{Both equations}~a+x=b, b+y=a~\text{have answers in }~\mathbb{Q}\}=\mathbb{Q}\times\mathbb{Q}$$

$$B=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{Both equations}~a\times x=b, b\times y=a~\text{have answers in}~\mathbb{Q}\}=(\mathbb{Q}\setminus \{0\})\times (\mathbb{Q}\setminus \{0\})\cup \{(0,0)\}$$

Now consider:

$$C=\{(a,b)\in\mathbb{Q}\times\mathbb{Q}~|~\text{All equations}~a^x=b, b^y=a, z^a=b, t^b=a~\text{have answers in}~\mathbb{Q}\}$$

$C$ is non-empty trivially because $(1,1)\in C$, is $C$ infinite?

Answer 1

$C$ is indeed infinite: it contains all pairs $(a,b)=(\frac1n,\frac1n)$ for integer $n>0$; solutions are $x=y=1$ and $z=t=n^{-n}$.