I do not know the published proof, but here is an outline proof of the solvable case (which may be the same as the published one): we may take a finite solvable group $G$ of order divisible by $pqr$, and assume by induction that no proper subgroup of $G,$ and no proper homomorphic image of $G,$ has order divisible by $pqr.$ Hence we may suppose that $G$ has a minimal normal subgroup $M$ which is a Sylow $p$-subgroup. Arguing similarly, we may suppose that $G/M$ has a minimal normal Sylow $q$-subgroup $N/M,$ and that $G/N$ is cyclic of order $r.$ Let $Q$ be a Sylow $q$-subgroup of $G.$ Now $C_{G}(m) = M$ for each non-identity element of $M$ (or we are done). Hence $G$ is a Frobenius group with kernel $M$ (and complement a Hall $\{q,r\}$-subgroup of $G$ containing $Q$). Now it is well-known that (as $Q$ is elementary Abelian), we even have $Q$ of order $q$. Hence $G$ has a subgroup of order $qr$, which must be cyclic (as it is contained in a Frobenius complement), as was known to Burnside.

I am not sure what sort of generalization you had in mind to non-solvable groups. For example ${\rm PSL}(2,5)$ and ${\rm PSL}(2,7)$ are groups whose order has three prime divisors in which every element has prime power order.

Further to Frieder Ladisch's answer, I believe that it is possible to describe the structure of all finite groups in which every element has prime power order. This is a question of collecting together several characterization theorems- I may well be duplicating known arguments, though I am not sure whether these results have been collected together in one place in the literature, or in some group theory text. In particular, results of M. Suzuki are the main tools.

Let $G$ be a finite group which is not a $p$-group for any prime $p,$ but in which all elements have prime power order. Suppose first that $F(G) \neq 1.$ Then $G$ has an elementary abelian minimal normal $q$-subgroup $Q$ for some prime $q.$ Also, $C_{G}(Q) = F(G) = O_{q}(G).$ Now for any other prime divisor $p$ of $|G|$, the Sylow $p$-subgroups of $G$ are cyclic or generalized quaternion. Suppose first that $q$ is odd. If $[G:O_{q}(G)]$ is even, then by the Brauer-Suzuki theorem (if the Sylow $2$-subgroup of $G$ is generalized quaternion) we see that $G = O_{2^{\prime}}(G)C_{G}(t)$ for some involution $t.$ Since $C_{G}(t)$ is a $2$-group, we see that $O_{2^{\prime}}(G)$ is an Abelian $q$-group for the given prime $q$. On the other hand, if $q$ and $[G:O_{q}(G)]$ are both odd, then $G/O_{q}(G)$ is metacyclic, so has a normal Sylow $p$-subgroup its largest prime divisor $p.$ Then $G/O_{q}(G)$ must be a cyclic $p$-group by the earlier argument.

If we have $F(G) = 1,$ then $G$ has a unique component $L,$ which is simple. Furthermore, $L \leq G \leq {\rm Aut}(L),$ and $L$ is a simple CIT-group (the centralizer of every involution is a $2$-group). These were classified by M. Suzuki. Not all these have the property that every element has prime power order, but this reduces the problem to checking which possibilities for $L$ have this additional properties, and which subgroups of ${\rm Aut}(L)$ do.

Consider the case that $q =2$ and $F(G)$ a $2$-group. If there is a solvable normal subgroup $N$ of $G$ with $N > F(G)$, then $G/O_{2}(G)$ is easily checked to be metacyclic. If there is no such normal subgroup $N,$ then $G/O_{2}(G)$ has the same structure as described in the single component case, though it may be that not all possibilities for the component occur, since we also require that non-identity elements of odd order act fixed point freely on $O_{2}(G).$ Note that at least some cases occur here though: the semidirect product of ${\rm SL}(2,2^{n})$ with its natural module is one such example.