Let $\alpha$ be the angle such that $\cos\alpha=3/5,\sin\alpha=4/5$. Consider the points $P_n(\sin(2^n\alpha),\cos(2^n\alpha))$ for $n\geq 1$, all lying on the unit circle. By looking at the triangle formed by $(0,0),P_n,P_m$ we find that the distance between $P_n$ and $P_m$ is
which (trigonometry exercise) is rational because of our choice of $\alpha$. This gives a countable set as in question (since no three points on a circle are collinear).
To prove there is no larger one, consider any two points $P,Q$ in our set. For any pair of rational numbers $a,b$ there are at most two points in the plane at distance $a$ from $P$ and at distance $b$ from $Q$. Hence there can only be countably many other points in this set. Similar argument (starting with points $P,Q,R$ or so) shows that there are no uncountable such sets even in the Euclidean 3D space, nor are there in higher dimensional ones.
Let me remark that it is an old open question, posed in 1945 by Ulam (see here for some references) whether there is such a (countable) set of points with pairwise rational distances which is dense in the plane. It's a simple exercise to show that if there was one, then there would be one with no three points collinear as well.