Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z[t]$ be unbounded as n varies? 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!
 A: Let $F(x,y)$ be any degree 3 polynomial. So for example, you can take $F(x,y)=xy(x\pm y\pm 1)$. The original question asked about solutions to $F(x,y)=n$ with $x,y \in \textbf{Z}$, and the current question replaces this with $x,y \in \textbf{Z}[t]$. More generally, let $n\in\textbf{C}[t]$ be a nonconstant polynomial, let $I(n)$ be the number of solutions to $F(x,y)=n$ with $x,y \in \textbf{C}[t]$, and let $R(n)$ be the rank of the group of points on the elliptic curve $E_n : F(x,y)=n$ with $x,y\in\textbf{C}(t)$. 
Theorem: If $E_n$ is an elliptic curve and if the $j$-invariant of $E_n$ is not in $\textbf{C}$, then there is an absolute constant $K$ such that $I(n) \le K^{1+R(n)}$. 
This is a result of Marc Hindry and mine. (We actually only prove it for standard Weierstrass equations, but the proof is the same in general.) The reference is Invent. Math. 93 (1988), 419-450, although it also relies heavily on J. Reine Angew. Math. 378 (1987), 60-100. If $E_n$ has constant $j$, there's a similar statement, but one may need to untwist first. For example, taking $F(x,y)=xy(x+y)$, one needs to restrict to $n$'s that are cube-free. Also, of course, one has to avoid $F$ and $n$ values such that the curve $F=n$ has a singularity, since then $E_n$ is no longer an elliptic curve.
All of this doesn't answer the question, but it suggests that it will be hard to make $I(n)$ large, since it's not even known that $R(n)$ can be arbitrarily large.
A: The $\mathbb{Z}[t]$ example above with 3 (pairs of) positive solutions for $ab(a+b-1)=n$ has 5 in $\mathbb{Z}[u,v]/(u^2-2v^2+1)$ if $t=\frac{(2v+u)(u-v)}{2}$. This gives us a parametric family of curves in $ab(a+b-1)=n$ in $\mathbb{Z}$ with 5 pairs of positive integer points . We need $u>v$ so first cases are $(u,v,n)=(7,5,21420^2),(41,29,840456540^2),(99,70,32947602728040^2)$. In the first case $n=21420^2$ there is a further sixth pair of solutions but not for the next the next 4 cases.
Suppose that $u^2=2v^2-1$  and let $t=\frac{(2v+u)(u-v)}{2}$. Then we have the same three solutions as before for $ab(a+b-1)=(2t(t+1)(2t+1))^2$. There are also two more. (plus the same things with $a$ and $b$ swapped). These are as follows with the second and third being the new points:
$[{\small 1}, \frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2},\frac{(2v+u)(2v-u)(u+v)(u-v)uv}{2}]$
$[\frac{(u-v)^2(2v+u)u}{2},{\small (2v-u)(2v+u)v^2}, \frac{(2v-u)(u+v)^2u}{2}]$
$[\frac{(2v-u)^2(u+v)v}{2}, {\small(u-v)(u+v)u^2}, \frac{(u-v)(2v+u)^2v}{2}]$
$[\frac{(2v+u)^2(u-v)^2}{2}, \frac{(2v-u)^2(u+v)^2}{2}, { \small u^2v^2}]$
$[\frac{(2v+u)(u-v)uv}{2}, \frac{(2v-u)(u+v)uv}{2}, {\small (2v-u)(2v+u)(u-v)(u+v)}]$
note that $\frac{2v-u}{u-v},\frac{v}{u},\frac{2v+u}{v+u}$ are three successive convergents to $\sqrt{2}$. Of course these can be rewritten using the relation $u^2=2v^2-1$ and/or substitutions, for example the product $(2v-u)(2v+u)=1$.
In another direction is again 3 points but in $\mathbb{Z}[t,x]$
${\small [1+(4t+2)x, 2(t+1)(2t+1)(t+(4t^2+2t-1)x), 2t(2t+1)(t+1+(4t^2+6t-1)x)]}$
${\small [2t(t+(4t^2+2t-1)x), 2(t+1)(t+1+(4t^2+6t-1)x), (2t+1)^2(1+(4t+2)x)]}$
${\small [(2t+1)(t+(4t^2+2t-1)x), (2t+1)(t+1+(4t^2+6t-1)x), 4t(t+1)(1+(4t+2)x)]}$
For a majority of values for $t,x$ the resulting curve over $\mathbb{Z}$ has three positive points. Hower cases of 4 and 5 do happen quite frequently and even 6 sometimes. This suggests that more is there to be found. Replacing $x$ with $-x$ and then multiplying through by $-1$ gives curves $ab(a+b+1)=n$
The first example represents two iterations of a process:


*

*Find a family of examples in $\mathbb{Z}$ (say look at ones with at least 3 solutions, one being $[1,b,b]$, from those notice that a few have the other two triples very close to each other.

*Find a parametric form in $\mathbb{Z}[t]$.

*Put in integers for $t$ to get curves over $\mathbb{Z}$ with at least three points and look for cases with higher rank.

*Note that several have 5 solutions and prime factors familiar from convergents to $\sqrt{2}.$

*Find a parametric form of that.
The second example follows a similar process but this is long enough.
A: This is not an answer, just a long comment.
Out of opportunism I tried multiplying Aaron's record by a square. $(2t+3)^2$ turned somewhat lucky.
$n$ with 3 solutions: $2^{2} \cdot t^{2} \cdot (t + 1)^{2} \cdot (2t + 1)^{2} \cdot (2t + 3)^{2}$
Solutions:
$$\left[2^{2} \cdot t \cdot (t + 1), (t + 1) \cdot (2t + 1)^{2}, t \cdot (2t + 3)^{2}\right]$$
$$\left[2 \cdot (2t + 3) \cdot t^{2}, (2t + 1) \cdot (2t + 3), 2 \cdot (2t + 1) \cdot (t + 1)^{2}\right]$$
$$\left[2 \cdot t \cdot (t + 1) \cdot (2t + 1) \cdot (2t + 3), 1, 2 \cdot t \cdot (t + 1) \cdot (2t + 1) \cdot (2t + 3)\right]$$
Another $n$ with 3 solutions is $t^{2} \cdot (t + 2)^{2} \cdot (t + 1)^{4}$
Later, using a symbolic approach found these 5 $n$ with 3 solutions: $[t^{8} + 8t^{7} + 26t^{6} + 44t^{5} + 41t^{4} + 20t^{3} + 4t^{2}, 1024t^{8} + 4096t^{7} + 6528t^{6} + 5248t^{5} + 2212t^{4} + 456t^{3} + 36t^{2}, 256t^{8} + 1024t^{7} + 1664t^{6} + 1408t^{5} + 656t^{4} + 160t^{3} + 16t^{2}, 64t^{8} + 384t^{7} + 928t^{6} + 1152t^{5} + 772t^{4} + 264t^{3} + 36t^{2}, 64t^{8} + 384t^{7} + 928t^{6} + 1152t^{5} + 772t^{4} + 264t^{3} + 36t^{2}]$
