Can a totally real cubic with $q(0)=-1$, $q(1)=1$ divide $-z^a+(1-z)^u+(1-z)^jz^p$?

For $a,u,j,p \in \mathbb{Z}_{>0}$ put $T_{a,u,j,p}(z) \;=\; -z^{a} + (1-z)^{u} + (1-z)^{j}z^{p} \;\in\; \mathbb{Z}[z],$ so $T(0)=1$, $T(1)=-1$. Any monic cubic factor has $|q(0)|=|q(1)|=1$, i.e. its roots $\theta$ satisfy: $\theta$ and $1-\theta$ are both units. A monic cubic factor with a root in $(0,1)$ must have $q(0)=-q(1)=\pm1$; the case $q(0)=1$ gives Shanks' family $z^3-(t+3)z^2+tz+1$ and does occur. I ask about the other case, $q_c(z) = z^3+(1-c)z^2+cz-1, \qquad \operatorname{disc}(q_c)=c^4-10c^3+31c^2-30c-23,$ positive exactly for $c \le -1$ and $c \ge 6$.

Question. Can $q_c \mid T_{a,u,j,p}$ with $\operatorname{disc}(q_c)>0$ and $c \notin\{-1,6\}$?

Such factors do occur, but always with negative discriminant: over all $10^4$ tuples with exponents $\le 10$ there are $83$, all with $\operatorname{disc} \in \{-23,-31\}$, the smallest being $T_{1,1,1,2}=-(z^3-z^2+2z-1)$. These come in infinite families, because there the unit rank is $1$: with $\eta$ fundamental, $\theta=\eta^2$, $1-\theta=\eta^3$ and $\eta+\eta^5=\eta^2+\eta^3=1$, so the equivalent unit equation $\tau^u\theta^{-a}+\tau^j\theta^{p-a}=1$ becomes two linear conditions on $(a,u,j,p)$. If $q$ is totally real the rank is $2$ and Baker–Győry gives finiteness for each fixed $q$, but I see no uniformity in $c$.

Root geometry does not decide it: the required configuration is realizable by a totally real cubic of this shape, e.g. $z^3+2z^2-z-1$ ($c=-1$). One promising reformulation: with $\tilde T(z)=(1-z)^{\deg T}T(1/(1-z))$, every Shanks factor divides $\tilde T$ (immediate), no $q_c$ factor found does, and if $q_c \mid \tilde T$ were forced then its three distinct $\sigma$-conjugates, $\sigma(z)=1/(1-z)$, would all divide $T$, giving nine real roots — impossible, since $T$ has at most three.

Is there a known result on divisibility of such trinomials by exceptional-unit cubics?