Quartic polynomial taking infinitely many square rational values?

I was wondering whether the value of
$$P(x)=x^4-6x^3+9x^2-3x,$$
is a rational square for infinitely many rational values of $x$. Is there a general method to check this for a polynomial (in one variable)? If not, how would one go about figuring this out?

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If your quartic polynomial to be made a square,

$$F(x) = z^2$$

has an initial rational point, then there is a birational transformation that can transform this to an elliptic curve’s Weierstrass normal form. However, if you are in a rush to learn a simple method (known back to Fermat), then one way is this: Using any non-zero initial solution $x_0$, do the transformation,

$$x = y+x_0\tag1$$

For your curve, we have $x_0 = 1$ and I get,

$$F(y) = y^4-2y^3-3y^2+y+1$$

Assume it to be equal a square,

$$y^4-2y^3-3y^2+y+1 = (ay^2+by+c)^2\tag2$$

Expand, then collect powers of $y$ to get the form,

$$p_4y^4+p_3y^3+p_2y^2+p_1y+p_0 = 0\tag3$$

where the $p_i$ are polynomials in $a,b,c$. Then solve the system of three equations $p_2 = p_1 = p_0 = 0$ using the three unknowns $a,b,c$ to eliminate the $y^2, y^1, y^0$ terms. One ends up with,

$$105/64y^4+3/8y^3=0$$

Thus, $y =-8/35$ or,

$$x = y+x_0 = -8/35+1 = 27/35$$

and you have a new rational point $x_1 = 27/35$. Use this on $(1)$,

$$x = y+27/35$$

and repeat the procedure. Then,

$$x_2 = 3\times 4286835^2/ 37065988023371$$

Repeating it ad infinitum, one finds an infinite number of rational points $x_i$.