Seeking help extending Vieta-jumping to higher powers I am trying to prove the following conjecture.
Conjecture. If $r > s \ge 1$ are relatively prime integers such that
\begin{equation}
  (r-s)^4-1 \equiv 0\!\pmod{4r^2s},  \tag{1}
\end{equation}
then $r-s = 1$ or $2r > 3s$.
A brute-force computer search has found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$.
Included below is the proof I have so far. I have attempted to extend the classic Vieta jumping technique to a cubic. My main question is: Having proved my conjecture for a cubic with three real roots, must I continue with the case where the cubic has two complex conjugate roots? Or is my claim valid, that fixing $k$ means I only have to consider the real roots?

Proof (possibly incomplete?).
Evidently the congruence is satisfied when $r-s=1$. Now assume $r-s > 1$, and write $g=r-s$. Since $r$, $s$, and $g^4-1$ are all positive, (1) now implies $g^4-1 = 4kr^2s$ for an integer $k \ge 1$. Hence
\begin{equation}
 g^4-1 = 4kr^2s = 4k(g+s)^2s = 4ks^3+8g ks^2 + 4g^2 k s,  \tag{$\star$}
\end{equation}
which we can rewrite as the cubic equation
\begin{align*}
 4ks^3 + 8g ks^2 + 4g^2 ks + (1-g^4) = 0.
\end{align*}
Fix $k$. Dividing through by the leading coefficient $4k$ and making the substitution $s \mapsto w$ yields
\begin{align}
 w^3 + 2g w^2 + g^2 w + \frac{1-g^4}{4k} = 0.
\end{align}
This cubic equation has three roots; call them $w_1, w_2, w_3$. Since $s$ is one root, fix $w_1=s$ without loss of generality. Now Vieta's formulas give
\begin{align}
 -2g = s + w_2 + w_3, &&
 g^2 = sw_2+ sw_3 + w_2w_3, &&
 \frac{g^4-1}{4k} = sw_2w_3.
\end{align}
Combining the first two relations implies
\begin{align*}
 4g^2 = (-2g)^2 = (s + w_2 + w_3)^2
  = s^2 + w_2^2 + w_3^2 + 2(sw_2+ sw_3 + w_2w_3)
  = s^2 + w_2^2 + w_3^2 + 2g^2,
\end{align*}
which gives
\begin{equation}
 2g^2-s^2 = w_2^2+w_3^2. \tag{2}
\end{equation}
A cubic function either has three real roots, or it has one real root and two nonreal complex conjugate roots. Since $w_1=s$ is real, $w_2$ and $w_3$ are either both real or they are nonreal complex conjugates. Since we have fixed $k$, and are seeking an integer solution with $1 < g = r-s$, we may take $w_2,w_3 \in \mathbb{R}$, where [at least] one of them is a second solution of ($\star$) for our fixed quotient $k$.
Now (2) implies $2g^2-s^2 = w_2^2+w_3^2 > 0$, yielding
\begin{equation*}
 s^2 < 2g^2 = 2(r-s)^2.
\end{equation*}
Therefore
\begin{equation*}
 \biggl(\frac{s}{r-s}\biggr)^{\!2} < 2 \qquad\implies\qquad \frac{r}{s} > \frac{1+\sqrt{2}}{\sqrt{2}} > \frac{3}{2},
\end{equation*}
and hence $2r > 3s$ as claimed.
 A: Take your equation $4ks^3 + 8g ks^2 + 4g^2 ks + (1-g^4) = 0$ and solve for $k$ to get $$k={\frac {{g}^{4}-1}{s (g+s)^2 }}.$$ First substitute $s=w$ in the equation and then substitue for $k$ to get (after dividing out $1-k^4$, and clearing the denominator) $${w}^{3}+2\,g{w}^{2}+{g}^{2}w-{g}^{2}s-2\,g{s}^{2}-{s}^{3}=0
$$ Since we know that $w=s$ is a root we have the courage to factor and end up with $$ \left( w-s \right)  \left({w}^{2}+w(s+2\,g)+ (g+s)^2
 \right) =0.
$$
Evidently, since $g,s \gt 0$,  the other two roots are complex.
This doesn't rule out solutions (obviously) since there are at least two. But it does mean this attempt to generalize the  Vieta-jumping technique to higher than degree two does not look promising. I don't know how classic it is. I learn from Wikipedia that around 20 years ago it solved a tricky Math Olympiad problem and in following years many problems have needed that technique. But perhaps it should be better known than it is.

I wondered if one could (in search of a contradiction) set $r=s+t$ with $2t \lt s$ and perhaps further let $s=(m-1)t+u$ with $1 \le u \lt t$ and appropriate things relatively prime. So $t^4-1$ is a multiple of $4(mt+u)^2(mt+t+u).$ It seems promising to me but I got bogged down. Note that $t \gt 4m^2(m+1)$ Also $t$ is odd, less than $s/2$ and such that $t^4 \equiv 1 \bmod s.$ This means $s$ can't be a prime or twice a prime. It is not hard to get $m^4-u^4 \equiv 0 \bmod mt+u$ and other such results but I don't see how to fit them all together.
