Is the following conjecture true?

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 so far found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$. [n.b. Noam Elkies confirmed the conjecture up to $r = 3 \cdot 2^{22} > 1.25 \cdot 10^7$ using gp.]

I posted a partial proof on MSE, but got no help despite several upvotes and a bounty offer (which has since expired).

The motivation for the proof is the application of Vieta jumping to equations greater than the second degree, primarily as a method of attacking Thue equations. So although any proof of the conjecture would be nice, a completion of my partial proof would be preferred.

Is the following conjecture true?

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 so far found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$. [n.b. Noam Elkies confirmed the conjecture up to $r = 3 \cdot 2^{22} > 1.25 \cdot 10^7$ using gp.]

I posted a partial proof on MSE, but got no help despite several upvotes and a bounty offer (which has since expired).

The motivation for the proof is the application of Vieta jumping to equations greater than the second degree, primarily as a method of attacking Thue equations. So although any proof of the conjecture would be nice, a completion of my partial proof would be preferred.

Is the following conjecture true?

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 so far found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$.

I posted a partial proof on MSE, but got no help despite several upvotes and a bounty offer (which has since expired).

The motivation for the proof is the application of Vieta jumping to equations greater than the second degree, primarily as a method of attacking Thue equations. So although any proof of the conjecture would be nice, a completion of my partial proof would be preferred.

Is the following conjecture true?

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 so far found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$. [n.b. Noam Elkies confirmed the conjecture up to $r = 3 \cdot 2^{22} > 1.25 \cdot 10^7$ using gp.]

I posted a partial proof on MSE, but got no help despite several upvotes and a bounty offer (which has since expired).

The motivation for the proof is the application of Vieta jumping to equations greater than the second degree, primarily as a method of attacking Thue equations. So although any proof of the conjecture would be nice, a completion of my partial proof would be preferred.