Are there infinitely many positive integer solutions to $(3+3k+l)^2=m\,(k\,l-k^3-1)$? I usually work in the field of differential geometry, but I have encountered the following problem in my research: Are there infinitely many positive integers $k,l,m\in\mathbb N^{>0}$ such that  $$(3+3k+l)^2=m\,(k\,l-k^3-1)\,?$$ Obviously, taking $l=k^2$
and $m=-(3+3k+l)^2$
gives infinitely many integer solutions, but  $m<0$ is negative. As a non-expert, I imagine that there is either a simple answer to this question, or the problem is not so simple to solve. Of course, I've played around with the equations a bit, but other than finding numerous examples, I haven't made any progress.
I would appreciate an existence or non-existence statement for infinitely many positive integer solutions, but also some hints that the problem is most likely hard to solve would help me.
Background: I am looking for certain integer representations of a surface group, and I can show that integer solutions to this diophantine equation actually give rise  to integer representations. The condition that $k,l,m$ are positive is equivalent to the condition that the corresponding representation is contained in a higher Teichmüller component (which is important for my differential geometric application).
 A: Here is another way maybe simpler as it reveals a certain Pell-Fermat equation. I don't know the relation with the other answer but i did many restrictions to get the following infinite sequence of solutions:
For $n\ge 1$, $$\begin{cases}q_n=-7q_{n-1}+24d_{n-1}\\d_n=2q_{n-1}-7d_{n-1}\end{cases}$$
with $q_0=3$ and $d_0=1$, (solutions of $q^2-12d^2=-3).$
Set $j=2|d_n|$, $i=4|d_n|+|q_n|$,  $r=3(i+j)$, so that $m=r^2$, $k=ij-1$ and $x:=3+3k+l=\dfrac{mk+r|j^3-i^3|}{2}$. This implies the positive integer solutions $(m,k,l)$ are infinite.
The starting equation to solve in $ x $ over $\mathbb{N}$ was $x^2-mkx+m(k+1)^3$.
Its discriminant is a square $X^2=(r\cdot y)^2$ as
$\Delta=X^2=m^2k^2-4m(k+1)^3$, say $m=r^2$ is a perfect square, then $(rk-y)(rk+y)=4(k+1)^3$ and set $(k+1)^3=i^3j^3$, ($k=ij-1$).
$$\begin{cases}rk-y=2i^3\\rk+y=2j^3\end{cases}.$$
So $r=(i+j)(-1+\dfrac{i^2+j^2-1}{ij-1})$ and $y=j^3-i^3$.
Let $\dfrac{i^2+j^2-1}{ij-1}=s$ an integer say $s=4$.
Finally we solve in $i$, $i^2+j^2-1=4ij-4$, taking again its discriminant to be a perfect (even)  square $p=2q$, $j=2d$ we solve $p^2-12j^2=-12$ as $q^2-12d^2=-3$.
A: It does have infinitely many positive solutions. Here is just one such series.
Consider the following recurrence sequence:
$$u_0=1,\ u_1=2,\ u_{n+1} = 23 u_n - u_{n-1} - 4\qquad (n\geq 1).$$
Let $t,k$ be any two consecutive terms of this sequence, then setting $l:=k^2+t$ produces the following equality:
$$(3+3k+l)(t+1) = (k+26)(kl-k^3-1),$$
which gives solution $m:=\frac{(k+26)(3+3k+l)}{t+1}$ (which is an integer) to the original equation.

In fact, integrality of $m$ follows from the identity:
$$(u_{n+2}+1)(u_n+1) = (u_{n+1}+26)(u_{n+1}+1),$$
which can be verified from the recurrence for $u_n$.
In summary, the values $(k,l,m)$ in this solution series are given by
$$\begin{cases}
k = u_{n+1}, \\
l = u_{n+1}^2 + u_n, \\
m = (u_{n+2}+2)(u_{n+1}+2) + 24.
\end{cases}\qquad (n\in\mathbb{Z}_+)
$$

ADDED. I've added $u_n$ to the OEIS as sequence A350917. Together with 9 other similar recurrences it gives all solutions $k$ to $(tk-1)\mid (k+1)^4$, which are now listed in sequence A350916.
