Inspired by @PeterMueller, I believe I found a proof that $r = 3$.

Because of how this equation was obtained in the first place, I can assume $s \ge 2$ is even, and $r \ge s+1$ is odd. Writing $s=2v$ and $r=2v+2t+1$, substituting, and factoring yields \begin{align} 2v(v-2t-1)(2v+2t+1)^2 &= t(t+1)(2t^2+2t+1), \end{align} at which point a reverse substitution gives \begin{align} (v-2t-1)sr^2 &= t(t+1)(2t^2+2t+1). \qquad(\star) \end{align} For any hypothetical solution with $r > 3$, we have $3s > 2r$ (shown earlier). A quick calculation then yields $t < (r-2)/4$, which can be shown to contradict ($\star$).

Does that look right?

Thanks!
Kieren.