Here's an elaboration on R. van Dobben de Bruyn's answer.

Let $A$ be an abelian group. Define the group $\mathrm{Di}(A)$ as follows: (a) consider the direct product $A'=A\times C_2$, denoting $y$ the nontrivial element of $C_2$. (b) perform the semidirect product $A''=C_4\ltimes A$, with $\pm$-action. Denote by $z$ the element of order 2 of the acting $C_4$: it remains central in $A''$, and so does $y$. (c) Obtain $\mathrm{Di}(A)$ (generalized dicyclic group on $A$) by modding out $A''$ by the central subgroup of order 2 $\langle z^{-1}y\rangle$. Still denote by $y$ the image of $y$ in $\mathrm{Di}(A)$, it it central of order $2$.
Note that $\mathrm{Di}(A)/\langle y\rangle$ is the dihedral product $C_2\ltimes_\pm A$.

In $A''$, denoting by $t$ a generator of $C_4=\{1,t,z,t^{-1}\}$ and writing $A$ additively, we have $(t^{\pm},a)^2=(z,0)$, $(z,a)^2=(1,a)^2=(1,2a)$ for $a\in A'$. In particular, $\eta=(z,y)$ (which is killed in $\mathrm{Di}(A)$ is not a square. Hence the elements of order $\le 2$ in $\mathrm{Di}(A)$ are the images of elements of order $2$ in $A''$, which are the elements of the form $(1,a),(z,2a)$ with $2a=0$. In particular, if $A$ has no element of order $2$, these elements are $(1,0),(1,y),(z,0),(z,y)$, which in $\mathrm{Di}(A)$ is reduced to $\{1,y\}$. That is, if $A$ has no element of order $2$ then the only element of order $2$ in $\mathrm{Di}(A)$ is $y$.

Also this shows that all elements $(t^\pm,a)$ map to elements of order $4$ in $\mathrm{Di}(A)$, which therefore has infinitely many elements of order $4$ if $A$ is an arbitrary infinite abelian group.

Now the construction $A\mapsto\mathrm{Di}(A)$ is clearly functorial under group isomorphisms. Hence, if $\Gamma$ acts on $A$ by automorphisms, then it naturally acts on $\mathrm{Di}(A)$ by automorphisms: in steps: (a) extend in the trivial way the action to $A'=A\times C_2$, then (b) extend in the trivial way to $C_4\ltimes A'$ (acting trivially on $C_4$): this works because the $C_4$-action, by $\pm$, commutes with the $\Gamma$-action on $A'$; finally this action fixes $\eta=(z,y)$ and hence passes to the quotient to an action on $\mathrm{Di}(A)$, defining a semidirect product $\mathrm{Di}\rtimes\Gamma$.

From what's preceding, we immediately get: if $\Gamma$ has no element of order $2$, then the only element of order $2$ in $\mathrm{Di}(A)\rtimes\Gamma$ is $y$; if $\Gamma$ is infinite then it contains $\mathrm{Di}(A)$ hence has infinitely many elements of order $4$.

Finally we can choose $A=C_n^{(\Gamma)}=\bigoplus_{\gamma\in\Gamma}C_n$ for some odd $n>1$, and choose $\Gamma$ of finite odd exponent $q>1$. Here $n$ and $q$ are unrelated, can be chosen equal or not (they could be chosen even for the construction but then this will produce infinitely elements of order $2$). Then the resulting group $$G=\mathrm{Di}(C_n^{(\Gamma)})\rtimes \Gamma$$ works: it has a single element of order $2$, infinitely of order $4$, and has exponent dividing $nq$. Actually modding out by $\langle y\rangle$, the resulting group admits, as subgroup of order $2$, the standard wreath product $C_n\wr \Gamma$ (in particular, all elements of order $4$ in $G$ lie in the nontrivial coset of the unique subgroup of index $2$).

(In general— arbitrary abelian group $A$, arbitrary $\Gamma$-action on $A$, we always get this subquotient killing the center $\langle y\rangle$ and passing to a subgroup of index $2$, which yields the semidirect product $A\rtimes\Gamma$. For instance, if we want $G$ to have Kazhdan's Property T, the permutational action is hopeless, but possibly some choice of $\Gamma$-module works, namely we need $A\rtimes\Gamma$ to have Kazhdan's Property T.)