It's known all groups of order $p^{m}q^{n}$ and all groups of odd order are solvable (By Burnside theorem and Feit-Thompson theorem).
Let $G$ be a group of order $2^{m}\cdot p^{n}\cdot q^{t}$ where $p\neq 3$ and $q\neq 3$ are prime.
Does anyone knows why $G$ is solvable?
Let $G$ be an unsolvable group. Since $G$ is a finite group, then it has a chief series. Since $G$ is unsolvable it is easy to show that $G$ has series: $1 \unlhd N \lhd H \unlhd G$ such that $H/N$ is a non-abelian simple group or $H/N$ is a direct product of isomorphic non-abelian simple groups. Since $G$ has three prime divisors, then $H/N$ has three prime divisors too. Since $3\notin \pi (G)$, by the following Lemma, we can get a contradiction. Therefore $G$ is solvable group.
Lemma: If $G$ is a simple with three prime divisors, then $G$ is isomorphic to one of the following groups: $A_{5}$, $A_{6}$, $PSL(2,7)$, $PSL(2,8)$, $PSL(2,17)$, $PSL(3,3)$, $PSU(3,3)$ or $PSU(4,2)$.(Ref: Herzog, M, On finite simple groups of order divisible by three primes only, J. Algebra, 120 (10), (1968), 383-388.)
Similar to the above discussion we can prove that all finite groups of order $ 2^{m}\cdot 3^{n}\cdot p^{k}$, where $p$ is prime but not in {5,7,13,17} are solvable groups.
Notice that the order of every subgroup and every factor-group of a group whose order has your form also has the same form.
Now, you can, for example, use a classification of finite non-cyclic simple groups without elements of order 6 from Gordon, L. M. Finite simple groups with no elements of order six. Bull. Austral. Math. Soc. 17 (1977), no. 2, 235–246,
and notice that the orders of all these groups do not have your form.
Another way is to use Thompson's theorem about minimal non-solvable groups. See the last three parts of the paper here: Thompson, John G. Nonsolvable finite groups all of whose local subgroups are solvable. IV, V, VI. Pacific J. Math. 48 (1973), 511–592, ibid. 50 (1974), 215–297; ibid. 51(1974), 573–630.
