Let $G$ be a 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.

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.