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@joro: That doesn't seem to be proven. At MSE, Vepir searched up to $10^7$ and found a one parameter family which does not extend to $4$ bases, and a single sporadic example, but there is no proof these are the only $3$ digit solutions.
Maybe I'm missing an important assumption, but tell me why this doesn't work: Fix $d$ and $B$. The unit group in the real quadratic field you mention is generated by say $a+b\sqrt{d}>1$, so for a given $T$, we can find solutions $(a+b\sqrt{d})^{-n}<B^{1/2}T^{-1}$ for every large enough $n$. These converge to $0$, so the difference of two can be arbitrarily small, while $z$ should be constant since we've fixed everything it depends on. Do you want $|x^2-dy^2|\neq|u^2-dv^2|$, or even more, in different square classes?
What do you mean when you ask for two pairs of integers to bounded away from each other? Are you looking for $|x-y\sqrt{d}-(u-v\sqrt{d})|>z$, for all $u,v,x,y$ for some $z$ depending only on $B, T, k,$ and $d$? Is $k$ only here so you can increase $B$ and $T$ together within some uniform bounds, or are you using it for something else? Should the $B$ in the definition of $k$ also have a squareroot?
You can take $a=1$, $b=c=d=0$, and $e$ squarefree, or just with some prime factor that isn't repeated. Then by Eisenstein, the quartic $F(x,1)$ is irreducible, so the form $F$ is as well. You can have non-zero $J$ in the same congruence class as well, by taking $d$ to be any multiple of the specified prime.
Do you want irreducible forms $F$ with, for example, $(I,J)=(12e,0)$, or are you looking for a family of forms for any $I\equiv 0 (3)$, $J\equiv 0 (27)$?
