Can we predict when a polynomial can take more than one perfect square value? We consider only polynomial with integer coefficients. We know how to determine the first perfect square value but we don't know how to determine subsequent perfect square values except by testing every single value of the variable x.
As an example we have p(x) = 9 x^2 + 13x + 2. we have p(x)=64 for x=2. We can see that there are no other values of x < 2 for which p(x) takes a perfect square value by simply testing x=1 and x=0. We don't want to test every single value.
Given the value of the variable x that produced the perfect square value of the polynomial, is there a method or an algorithm or a test that can tell us that there is no other value smaller than x that will produce another perfect square value of the polynomial?
As an example where p(x) can take a perfect square value more than once, we have p(x) = 9x^2 + 13x + 1. p(x)=121 for x=3 and p(x)=1 for x=0.
 A: I assume you're looking for integer values of $x$.  
If, as in your examples, your polynomial $p$ is quadratic, it's quite easy to determine the integer solutions to $p(x) = y^2$.  Look up "Pell's equation".
For polynomials of higher degree, the question is much more difficult. If the curve $p(x) - y^2$ has genus $> 1$, Faltings's theorem says there are only finitely many rational points, but I don't know if there are effective bounds.
A: When in equation square:
$$y^2=a^2x^2+bx+c$$
This is equivalent to the representation of a number as the difference of squares. Means decisions not endless. To start with $s=1$ and finish when $x$ is less than 2.
And use the formula.
$$y=\frac{a}{s}\left(c-\left(\frac{b+s}{2a}\right)^2\right)+\frac{b+s}{2a}$$
$$x=\frac{1}{s}\left(c-\left(\frac{b+s}{2a}\right)^2\right)$$
A: The problem is rather old and simple. Solved always standard. In Your case boils down to the need to solve equations of the form.
$$y^2=ax^2+bx+c$$
You can see the discussion here.  https://mathoverflow.net/questions/31118/integer-polynomials-taking-square-values
Although they persistently looking for the sequence, but for the case when:
$$y^2=ax^2+bx+q^2$$
The solution can be written quite simply.  To do this, use the following equation Pell.
$$p^2-as^2=1$$
Using the solutions of the Pell equation solution can be written as :
$$x=(2qp+bs)s$$
$$y=qp^2+bps+aqs^2$$
In order to find all solutions of this equation Pell often use the exponentiation of the sum with the root of the coefficient. It is difficult and does not make sense. It is better to make the sequence knowing the first solution of the Pell equation $(p_0 ; s_0 )$.  Then using any solution $(p_1 ; s_1 )$ can be found the following.  To do this, use the formula.  
$$p_2=p_0p_1+ay_0y_1$$
$$y_2=y_0p_1+p_0y_1$$
To find the first solution of the Pell equation easily. This is a standard procedure. Need to decompose the root factor in the continued fraction.  The difficulty will be when $c$ is not a square.
There are several formulas you can use. http://www.artofproblemsolving.com/blog/103509
http://www.artofproblemsolving.com/blog/101140
The problem still boils down to any of the Pell equation. For this purpose it is necessary to make the replacement of the $x\longrightarrow(x+r)$ or such $y\longrightarrow(y+r)$ . Further standard to seek solutions to the Pell equation. 
A: To explore the equation $$y^2=9x^2+13x+2$$ first multiply by $36=4\times 9$ to obtain $$(6y)^2=4\cdot(9x)^2+4\cdot 13\cdot9x+72=(18x+13)^2-97$$ so that $$(18x+13)^2-(6y)^2=(18x+6y+13)(18x-6y+13)=97$$
For integer solutions you have that $97$ is prime, so the factors are $1$ and $97$ or $-1$ and $-97$. Either way the difference between the factors is $12y=\pm96$ so that $y=\pm8$ and $x$ must be a solution of $9x^2+13x+2=64$. From there you have:
$$(9x+31)(x-2)=9x^2+13x-62$$
For $y^2=a^2x^2+bx+c$ multiply by $4a^2$ and complete the square to get $$(2ay)^2=(2a^2x+b)^2+4a^2c-b^2$$and proceed similarly. You have to test all factorisations at the end.
