Given five rational points $A,B,C,D,E$, we can find a sixth rational point as follows:
- Let $G$ be the intersction of $AB$ and $DE$.
- Let $H$ be a rational point on $CD$.
- The line $AH$ intersects the conic in another point $F$
- Let $K$ be the intersection of $GH$ and $EF$.
Now $G=AB\cap DE$, $H=CD\cap FA$, $K=EF\cap BC$ are on the Pascal line of the hexagon $ABCDEF$.
As $A,B,D,E$ are rational, so is $G=AB\cap DE$.
As $A,G,B,C$ are rational, $K=AG\cap BC$ is rational.
As $A,H,E,K$ are rational, $F=AH\cap EK$ is rational.
So in order to show that there are infinitely many rational points $F$, it suffices to show that the above construction is possible in infinitely many ways.
Step 1 is clear. Step 2 is possible in infinitely many ways because $CD$ is a rational line and contains infinitely many rational points.
For step 3, first note that $A$ is not on $BC$, hence the infinitely many lines $AH$ for different choices of $H$ are pairwise different.Also, there are at most two lines through $A$ that do not intersect the conic again. Hence step 3 still works for infinitely many $H$. Step 4 is "deterministic" again; note that in case of parallel lines, $K=\infty$ is still counted as rational.