No. The set of primes can be whatever you want (added: within reason! As Benjamin Steinberg points out, it can in fact be any recursively enumerable set of primes).
First, note that for infinitely presented groups, the torsion can be whatever you like: the torsion in the group
$*_i \mathbb{Z}/p_i$
is precisely the set of primes $p_i$, by standard facts about free products. (Added: As long as the set of $p_i$ is recursively enumerable, then this group admits a recursive presentation on a countable set of generators.)
By Higman's Embedding Theorem, the above group can be embedded in a finitely presented group. More subtlely, this embedding doesn't introduce any new torsion---see, for instance, Theorem 2.5 of this preprint of Chiodo.
Clarification: Higman's Embedding Theorem is commonly stated as only applying to finitely generated countable groups. In fact, an old construction of Higman, Neumann and Neumann shows how to embed a countably generated group into a 2-generated group; if the countably generated group is recursively presented, then the 2-generated group can be taken to be recursively presented as well.
Further update: Following Benjamin Steinberg's comments below answer and an argument in another paper of Chiodo (see also Francois G. Dorais's comments), I think we have a very exciting characterization of the sets of primes that can occur as torsion in finitely presented groups. (For the solvable-word-problem case, one also needs a theorem of Clapham, which says that the Higman Embedding can be made to preserve solvability of the word problem.)
Very exciting theorem:
Let $P$ be a set of primes.
- $P$ occurs as the torsion in some finitely presented group if and only if $P$ is $\Sigma^0_2$.
- $P$ occurs as the torsion in some finitely presented group with solvable word problem if and only if $P$ is recursively enumerable.