Assume $q \not| p^i - 1$ for $1 \leq i \leq n - 1$. Prove that $G$ is solvable.
Since if $G$ has a solvable normal subgroup $N$ such that $G/N$ is solvable, and if $r$ is prime, every $r$-group is solvable, we know that if $G$ has a normal Syl$_p$ group or a normal Syl$_q$ group then $G$ is solvable.
Suppose $G$ does not have a normal Syl$_p$ group or a normal Syl$_q$ group. Then $G$ must have $q$ many Syl$_p$ subgroups, and since $q \not| p^i - 1$ for $1 \leq i \leq n - 1$, and the number of Syl$_q$ subgroups $= 1$ mod $q$, we get that the number of Syl$_q$ subgroups is $p^n$
I'm not really sure what to do at this point but my idea was if I can show that $G$ has a normal non trivial $p$ subgroup $N$, then $G/N$ must have a normal sylow$_q$ subgroup and so by repeating my argument in the first paragraph I am done. But I cant figure out how to get such a group.
I know $p$-groups can't have trivial center so I am trying to think of a way to use that idea to get such a group. Another idea I had was if $P$ is a syl$_p$ group then since $|G/P| = q$. we know there exists a normal subgroup $K$ in $G$, contained in $P$, such that $|G/K|$ divides $q!$. So if I can show that $k$ is non trivial then $K$ is my non trivial $p$-group.
By Burnside's theorem, every group of order $p^mq^n$ where $p,q$ are prime,is solvable.
In your particular case, by the Sylow theorem the number of Sylow $q$-subgroups must divide $p^n$ and be congruent to 1 mod $q$. By your assumption this number is 1 or $p^n$. Suppose it is 1. Hence the Sylow $q$-subgroup $Q$ is unique and has $q$ elements whence is cyclic. Hence the group is an extension of $Q$ by $G/Q$ of order $p^n$. The $p$-group $G/Q$ is nilpotent, whence solvable. Therefore $G$ is solvable.Now suppose it is $p^n$. All Sylow $q$-subroups pairwise intersect at $1$ and have $q$ elements. Hence the group has $p^n(q-1)=p^nq-p^n$ elements of order $q$. Therefore there can be only one Sylow $p$-subgroup $P$ of order $p^n$. Hence $P$ is normal, nilpotent, and $G/P$ is cyclic. So $G$ is solvable.