Find all rational solutions of the equation
$$3x^2-5y^2=3$$
After a little bit of calculations I have managed to find a single pair $(4,3)$. I have heard about something called rational points on curves (e.g. circle).
Is there some "general algorithm" for solving this kind of problems?
Any rational solution $(x,y)$ must satisfy $$y-3=m(x-4)$$ for some rational number $m$. Now, write $3x^2-5y^2=3$ as $$3(x-4)(x+4)-5(y-3)(y+3)=0\,.$$ Thus, for a solution $(x,y)\neq (4,3)$, we have $$3(x+4)-5m(y+3)=0\,.$$ Plugging in $y=3+m(x-4)$, we get $$-\left(5m^2-3\right)x+\left(20m^2-30m+12\right)=3(x+4)-5m\big(6+m(x-4)\big)=0\,.$$ That is, $$x=\frac{20m^2-30m+12}{5m^2-3}\,.$$ Then, you can solve for $y$, using $y=3+m(x-4)$.
Once you have the first point you can do it (for conics). If there is another point $(x,y)$ on the curve with rational coordinates, then the equation of the line through $(x,y)$ and $(4,3)$ obviously has rational coefficients. Conversely, a line with rational slope through (4,3) intersects the curve in another point with rational coordinates (unless it happens to be the tangent.)
Look at these notes for an example with a circle.
