i figured it out:
let G act on the sylow p-subgroups by conjugation.
this gives a homomorphism of G into S2p+1.
since G is simple, and the action is transitive, the kernel of this action must be trivial.
therefore |G| divides (2p+1)!.
now we can ask, what is the highest possible power of p that divides (2p+1)! ?
well, the only integers 2 ≤ n ≤ 2p+1 that p divides are p and 2p. this means that the highest possible power of p that divides (2p+1)! is p2.
thus the sylow p-subgroups of G have either order p, or p2, and are abelian in either case.