Nontrivial rational solution of $x^3-3x^2+3x+1=y^2$ So, I'm looking for a rational solution of $$x^3-3x^2+3x+1=y^2$$ that is different from $(0,\pm 1)$. I tried to substitute $y=xt+1, t\in\mathbb Q$, getting following equation $$
x(x^2-(t^2+3)x+(-2t+3))=0$$ with the discriminant of the quadratic term $t^4+6t^2+8t-3$. The next step would be to find a $t$ for that  this discriminant is a square of a rational number. Could you give me a tip how to do it or propose another solution?

As hinted in the comments, we may consider the equivalent equation
$$y^2 = x^3 + 2$$
This is an elliptic curve. Finding rational points is an active domain in research and many books have been written, e.g. Silverman's Rational Points on Elliptic Curves.
Now, you already have some rational points on your curve: $(-1,-1)$ and $(-1,1)$. We can construct new rational points on the elliptic curve by taking the tangent line through one of these points, and calculating the intersection with the elliptic curve.
The tangent line to $(-1,1)$ is
$$y = \frac{3}{2}(x+1) + 1 = \frac{3}{2}x+\frac{5}{2}$$
Now, taking this $y$ in the original equation, we get:
$$\left(\frac{3}{2}x+\frac{5}{2}\right)^2 = x^3 + 2$$
Solving this cubic for $x$, we obtain two solutions: $x = -1$ and $x = \frac{17}{4}$. You can then calculate $y = \pm \sqrt{\left(\frac{17}{4}\right)^3+2} = \pm \frac{71}{8}$.
So, two more points on your curve are $\left(\frac{17}{4},\frac{71}{8}\right)$ and $\left(\frac{17}{4},-\frac{71}{8}\right)$.
You can then repeat this with the new solution(s) to find more of them, e.g.
$$(x,y) = \left(\frac{66113}{80656},\frac{36583777}{22906304}\right)$$
