I don't understand why we can foud the nilpotent operator?
Let S, T ∈ B(X) be invertible operators, Assume that there exists x ∈ X such that $S^{-1}Tx$ and x are linearly independent. Then we can find a nilpotent R of rank one satisfying $Rx = −S^{-1}Tx$
For convenience, say $u = -S^{-1}Tx$. You know $u$ is a nonzero vector in the range of $R$, and since the rank is $1$ this and its scalar multiples are the only ones. So $R: v \mapsto \phi(v) u$ where $\phi$ is a nonzero bounded linear functional on $X$, with $\phi(x) = 1$ to make $R(x) = u$. On the other hand, you want $R$ to be nilpotent so you'd better have $R(u) = 0$, i.e. $\phi(u) = 0$. Use Hahn-Banach to prove the existence of such a $\phi$ and you're done.
But I hope you're not too particular about that "find". Hahn-Banach uses the Axiom of Choice, and there may not be a guarantee that you can actually construct such a $\phi$.