Is there any possible way to solve this equation with two different variables with different degrees?

Question

$$649 + 96y = k^{2}$$ Also, $(y)$ or $(y + 1)$ or $(y-1)$ must be a perfect square. $y$ and $k$ are both natural numbers.

Answer 1

I asked Alpha to solve $649+96x^2=y^2$ over the integers, then replaced $649$ with $553$ and $745$. Only $745$ gave solutions. The lowest is $x=8,y=83$ and there are a few families. It clearly used the continued fraction approach to find them. The usual approach to show there are no solutions is to find a clever modulus to show the equation is impossible.