I have a polynomial $f(x)=a^2x^2+bx+c\in\mathbb Z[x]$ with $f(x)$ not a constant times a square and $abc\neq0$ and I want to know how many $x$ between $-a$ and $a$ the polynomial is a perfect square. Since leading coefficient is a perfect square by this reference the polynomial only has finitely many squares.

How many squares can be achieved by $f(x)$ with $x\in(-a,a)\cap\mathbb Z$ if $|a|^6,|b|^4,|c|^3$ are roughly of similar size? Gap between squares is squareroot in size and $(x-y)|(f(x)-f(y))$ makes me think there is at most only $1$ square with $x\in(-a,a)\cap\mathbb Z$.

Is there a way to test in polynomial time the parity of number of square $f(x)$ with $x\in(-t,t)\cap\mathbb Z$ at any given $t<a$ or find at least one such square?

Is there any good reference on the topic? It suffices to compute $\sum_{x=-t}^t\omega(f(x))\bmod 2$ where $\omega(f(x))$ is number of divisors of $f(x)$.