Integer solutions of a monic polynomial

I'm trying to find integer solutions to the equation $x^3 - k^2 x +(k-1) = 0$ where $kx$ is an integer and $k > 1$. Also if there is a way to find integral solutions to the same equation if $k$ is a rational such that $k>1$.

Claim:$\;$The only pairs $(x,k)$, where $x$ is an integer and $k$ is rational, satisfying the equation $$ x^3-k^2x+(k-1)=0 $$ are the pairs listed in the following table \begin{array} {c|c|c|c|} x&-1&-1&0&1&1\\ \hline k&-2&1&1&0&1\\ \end{array} Proof:

The cases $x=-1,\;x=0,\;x=1$ are easily resolved, and yield the solutions from the above table.

Next suppose $(x,k)$ is a valid solution with $|x| > 1$.

Regarding the given equation as a quadratic in the variable $k$, it follows that the discriminant $$ 4x^4-4x+1 $$ must be a perfect square.

But for $|x| > 1$ we have $$ (2x^2-1)^2 < 4x^4-4x+1 < (2x^2+1)^2 $$ so the only way for $4x^4-4x+1$ to be a perfect square is to have $$ 4x^4-4x+1 = (2x^2)^2 $$ but that would yield $x={\large{\frac{1}{4}}}$,$\;$contradiction.