The choice of having $c^2$ on the right hand side is irrelevant, any nonzero number gives the same conclusions.
Let $ g = \gcd(a,b). $ By unique factorization, with $ \gcd \left( \frac{a}{g}, \frac{b}{g} \right) = 1, $ the product being a square gives $ a = g \alpha^2, \; \; b = g \beta^2, $ and let us take $g, \alpha, \beta > 0.$ So your equation becomes $ c^2 = a x^2 - b y^2 = g (\alpha^2 x^2 - \beta^2 y^2) = g (\alpha x - \beta y) (\alpha x + \beta y). $ Now, either $\alpha x, \; \beta y$ have the same sign or opposite. With $c \neq 0$ we get get one factor at least $1$ in absolute value, so then $ |\alpha x| + | \beta y| \leq \frac{c^2}{g}, $ so $ | x| \leq \frac{c^2}{g \alpha} $ and $ | y| \leq \frac{c^2}{g \beta}, $ giving finiteness of the set of solutions.
MEANWHILE, if $ab$ is not a perfect square, there are infinitely many solutions to the Pell equation $ u^2 - a b v^2 = 1. $ This makes infinitely many different solutions if there are any, because $ a (ux + b v y)^2 - b (avx + uy)^2 = a x^2 - b y^2. $