What other prime numbers have been ruled out as counterexamples to the Feit-Thompson conjecture?
First, note that $(3^q-1)(q-1)>2(q^3-1)$ for every $q>3$. This is because $3$ is the point where the two lines meet, and $3^q$ grows faster than $q^3$.
Now, we simply let $q$ and $n$ be prime and natural (respectively) such that $n(\frac{3^q-1}{2})=\frac{q^3-1}{q-1}$.
Then, $n(3^q-1)=2(\frac{q^3-1}{q-1})$, and thus $n(3^q-1)(q-1)=2(q^3-1)$. This means that $q\leq3$. Of course, because the final equation also holds for both $2$ and $3$, this means that there are no numbers greater than 3 where this is true. So, there are no counterexamples.