We've recently seen this question: Can the number of solutions $ab(a+b+1)=n$ for $a,b,n \in \mathbb{Z}$ be unbounded as $n$ varies? It appears initially plausible that the answer is yes, but evidently there are good reasons to believe that the answer might be no. The same question and lack of certainty applies equally to $$ab(a+b-1)=n.$$

I wonder about the analogous situation with $\mathbb{Z}$ replaced by $\mathbb{Z}[t]$

Is there a known or conjectured upper bound for the number of solutions $ab(a+b-1)=n$ for $a,b,n \in \mathbb{Z}[t]$ as $n$ varies? What is the best known lower bound?

Meta-reasoning tells us that the number of solutions for this $\mathbb{Z}[t]$ problem is not know or suspected to be unbounded because that would supply us with parametric families of solutions to the problem in $\mathbb{Z}$.

I picked the $-$ version because I was able to find this nice example:

$$[a,b,a+b-1]=[1,2t(t+1)(2t+1),2t(t+1)(2t+1)],[2t^2,2(t+1)^2,(2t+1)^2]$$ and $$[t(2t+1),(t+1)(2t+1),4t(t+1)]$$ are three solutions to $$ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2.$$ That counts as 3 solutions. (or 6 or 18 if we take $[b,a,a+b-1]$ and $[1-a-b,a,-b]$ as different, though why bother?). Looking for integer solutions to $ab(a+b-1)=\left(2t(t+1)(2t+1)\right)^2\left(2t(t+1)(2t+1)\right)^2$ most often leads to just the three anticipated solutions but sometimes there are more. Including 6 for $t=55$ and 9 for $t=175$.

It seems reasonable to require that the leading coefficients be positive and that $n=n(t)$ not be an integer (I needed to say that before someone else did!).

The more general question arises as well. But I like concrete examples and don't want to venture a version of minimal Weierstrass form (feel free to enlighten me however)

I've further extended the previous update to this question and made it an answer. Please provide a better one!