Is there an infinite simple group with no element of order $2$?

Yes, because there exist torsion-free infinite simple groups (i.e. simple groups having no elements of finite order).

Here you can read a bit about one type of example of such groups.

Another 'yes' answer can be found through the Tarski Monster groups.

A group $G$ is a Tarski Monster $p$-group for some prime $p$ if $G$ is infinite, finitely generated, and every proper non-trivial subgroup of $G$ is cyclic of prime order. Such a group is necessarily simple (see below). Ol'shanskii proved that these existed for all "$p$ large enough" ($p>10^{75}$ or something similar) in a series of papers in the late 70s-early 80s. He published a book which contains the proof in '89 (Russian)/'91 (English), called The geometry of defining relations in groups.

Mark Sapir and others have done work recently on Monster groups in general. Ol'shanskii's construction is infinitely presented (but two-generated), and I believe it is still an open problem as to whether there exist finitely presented Tarski Monsters.

Below: Let $T$ be a Tarski Monster group. Then we prove that $T$ is simple.

Assume $T$ is not simple. Thus, there exists some $K$ such that $\langle 1\rangle\neq K\lhd T$. By the definition of $T$, $K=\langle x\rangle\cong C_p$, and as $K$ is normal we have that for arbitrary $g\in T$, $g$ acts on $K$ by conjugation. Thus, $\langle g\rangle\rightarrow \operatorname{Aut}(C_p)$. However, $\langle g\rangle\cong C_p$ while $|\operatorname{Aut}(C_p)|=\varphi(p)=p-1$ (in fact, it is well known that $\operatorname{Aut}(C_p)=C_{p-1}$) so the action of $g$ on $K$ is trivial. That is, $gxg^{-1}=x$ for all $g\in G$.

Now, let $g\in G\setminus K$ be arbitrary. Then either $\langle x, g\rangle=T$ or $\langle x, g\rangle\cong C_p$. However, $x$ and $g$ commute while $T$ is non-abelian so $\langle x, g\rangle\neq T$. On the other hand, $C_p=\langle x\rangle\lneq \langle x, g\rangle$ so $\langle x, g\rangle\not\cong C_p$. This is a contradiction. Thus, $T$ is simple.