Show that for $u, v \in \mathbb{Z} $ there are only a finite number of $a, b \in \mathbb{Z} $ such that $(ab)^2-ua-vb$ is a square.

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This is a generalization of my question Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer?

Show that for $u, v \in \mathbb{Z}^+ $ there are only a finite number of $a, b \in \mathbb{Z}^+ $ such that $(ab)^2-ua-vb$ is a square.

(I.e., all variables are positive.)

Note that there are two cases (at least in my analysis):

(1) $a$ and $b$ are sufficiently large. In this case, there are upper bounds on $a$ and $b$.

(2) $a$ or $b$ is small enough so that the bounds in case (1) do not apply. In this case it has to be shown that for any $a$ there are only a finite number of $b$ so that the expression is a square (and similarly for $b$).