I am trying to find all integral points on a certain family of elliptic curves using elementary methods, and it all boils down to division polynomials. In particular, define $\psi_0=0; \psi_1=1; \psi_2=2,\psi_3=-1,\psi_4=-36,$ and for $m\ge 3$ $$\psi_{2m}=\frac{1}{2}\psi_m(\psi_{m+2}\psi_{m-1}^2-\psi_{m-2}\psi_{m+1}^2),$$ $$\psi_{2m+1}=\psi_{m+2}\psi_{m}^3-\psi_{m-1}\psi_{m+1}^3.$$ One can prove that any two consecutive terms in the sequence are coprime. The question is, can we prove that $(\psi_m)_{m\ge 3}$ is an increasing sequence, or even the weaker statement that $\psi_m=\pm1$ if and only if $m\in\{1,3\}?$ The later statement must be correct, as the two values correspond to the only four integral points on the elliptic curve $$y^2=x^3+x+1.$$ Height functions can give the following inequality: $$275.4350088e^{0.476223106404866 n^2}\ge \max\{|\psi_{n-1}||\psi_{n+1}|,|\psi_n|^2\}\ge \frac{1}{160.068285}e^{0.476223106404866 n^2},$$ but it is not clear how we can use it.
Also, it seems likely that this can be extended to a general positive integer $t$: define $\psi_0=0; \psi_1=1; \psi_2=2t,\psi_3=-1,\psi_4=-4t(8t^4+1),$ and for $m\ge 3:$ $$\psi_{2m}=\frac{1}{2t}\psi_m(\psi_{m+2}\psi_{m-1}^2-\psi_{m-2}\psi_{m+1}^2),$$ $$\psi_{2m+1}=\psi_{m+2}\psi_{m}^3-\psi_{m-1}\psi_{m+1}^3.$$ How to find all $m$ such that $\psi_m=\pm 1$?