If $x$ and $b$ are positive integers and it is also given that the expression $x^2 + \dfrac{4x}{b^2} - 4$ is a perfect square. what is a good method to analyse the expression with the knowledge that it is a perfect square and $a$ and $b$ are positive integers, so that ultimately we can try to get some kind of relation or constraint between $a$ and $b$? One relation I could think was $(\dfrac{4x}{b^2}-4) = 0$, so that expression would reduce to $x^2$, which is a perfect square. but is it the only possibility and would like to know other possibilities if any?
$$\textbf{ Here is the attempt I made, please point out the faulty logic.}$$
A perfect square expression $(x+a)^2$ is expanded as $ f = x^2 + 2ax + a^2$ and $ f' = 2x + 2a $.
Further $ (f - af') = x^2 - a^2 $ and by gcd property $(x + a)$ is a factor of $ (f - af') $.
Using the above, if $g = x^2 + \dfrac{4x}{b^2} - 4$ ,
then $ (g - \dfrac{2}{b^2}g') = x^2 - \dfrac{8}{b^2} - 4$.
The expression $ (g - \dfrac{2}{b^2}g')$ should have a factor equivalent to $(x + a)$, if $g = x^2 + \dfrac{4x}{b^2} - 4$ is a perfect square.
This would mean that the term $( \dfrac{8}{b^4} + 4)$ in $ (g - \dfrac{2}{b^2}g')$ equates to $a^2$ term in $ (f - af') $.
Therefore the term $ (\dfrac{8}{b^4} + 4)$ must be perfect square and given that b is a positive integer.
For $b=1$, $ (\dfrac{8}{b^4} + 4)$ evaluates to 12, which is not a perfect square.
For $b=2$, $ (\dfrac{8}{b^4} + 4)$ evaluates to 6, which is not a perfect square.
For all other b, $ (\dfrac{8}{b^4} + 4)$ are not even integers.
This contradicts $ (\dfrac{8}{b^4} + 4)$ being a perfect square and in turn establishes that there is no $(x+a)$ type factor in $ (g - \dfrac{2}{b^2}g')$. So this means that for the expression $(x^2 + \dfrac{4x}{b^2} - 4)$ to be a perfect square , there is the only possibility of it reducing to $x^2$. This means $(\dfrac{4x}{b^2} - 4) = 0$.
Kindly point out any flaws in the proof.