For which integers $a,b$ does $ab-1$ divide $a^3+1$? A problem I wasn't able to solve:

For which values of $a,b\in\mathbb{Z}$ does $ab-1$ divide $a^3+1$?

I am looking for every possible solution. Some of them are trivial, like $a=0,b=0$ or $(a,b)\in\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,5),(5,3)\}.$
You may notice it is very similar to the famous 1988 IMO problem #6, and I bet Vieta jumping is the key, but an elliptic curve seems to be involved. You may also notice that $(ab-1)\mid (a^3+1)$ implies $(ab-1)\mid (a^3 b^3+b^3)$, hence $(ab-1)\mid (a^3+1)$ is equivalent to $(ab-1)\mid (b^3+1)$.

In order to avoid trivial cases, we may assume $|a|>1$ and $|b|>1$ without loss of generality. Given
$$ a^3+1 = (ab-1)\cdot k \tag{1}$$
we must have $k\equiv -1\pmod{a}$, i.e. $k=(ac-1)$. The previous identity then becomes:
$$ a^2-(bc)a+(b+c)=0 \tag{2}$$
hence every solution $(a,b)$ is associated with other solutions $\left(a,\frac{a^2+b}{ab-1}\right),\left(a,\frac{b^2+a}{ab-1}\right)$. 
$b^2 c^2-4(b+c)$ has to be a square: that obviously cannot happen if $b^2 c^2-4(b+c)<b^2 c^2$ and $b^2 c^2-4(b+c)>(bc-1)^2$: that observation leads to the fact that the only solutions in $\mathbb{N}$ are the ones listed above, but what about the other solutions in $\mathbb{Z}$?
 A: A solution can be obtained in a more direct manner than Vieta jumping.
First, if $ab-1\mid a^3+1$, then $ab-1\mid a^2+b$ since $b\cdot(a^3+1)=a^2\cdot(ab-1)+a^2+b$. Similarly, $ab-1\mid b^2+a$ and $ab-1\mid b^3+1$. So, $ab-1$ divides either all or none of $a^3+1$, $a^2+b$, $b^2+a$, and $b^3+1$.
In particular, as has already been noted, if $(a,b)$ is a solution, then so is $(b,a)$, so we may restrict ourselves to solutions with $|a|\le|b|$. For small, fixed $a$, we may simply check all $b=(k+1)/a$ for integers $k\mid a^3+1$. This gives us the following solutions with $|b|\ge|a|$, assuming I haven't missed any:


*

*$a=0$ and any $b$

*$a=1$ and $b=-1,2,3$

*$a=-1$ and any $b$

*$a=2$ and $b=-4,2,5$

*$a=-2$ and $b=-4,3$

*$a=3$ and $b=-9,5$

*$a=-3$ and $b=-9,4$


In the following, we may therefore assume $4\le|a|\le|b|$.
If $ab-1\mid a^2+b$, either $b=-a^2$, which is always a solution, or $|a^2+b|\ge|ab-1|$. The latter, we split into two cases based on the sign of $a^2+b$.
If $a^2+b>0$, we get $a^2+b\ge|ab-1|$ from which follows that $|a|^2+|b|\ge|a|\cdot|b|-1$. Still assuming $|b|\ge|a|$, this gives
$$
\bigl(|b|-|a|-1\bigr)\bigl(|a|-1\bigr)=|a|\cdot|b|-|a|^2-|b|+1\le2
$$
which implies that $|b|$ is either $|a|$ or $|a|+1$ when $|a|>3$. We can then check the alternatives $b=a$, $b=-a$, $b=a\pm 1$, $-b=a\pm 1$ (with $\pm$ depending on sign of $a$) to verify these give no additional solutions.
Similarly, if $a^2+b<0$, we have $-b>a^2$, which with $|b|\ge|a|\ge4$ makes
$$|ab-1|\ge|a|\cdot|b|-1>|b|=-b>-b-a^2>0$$
for which there is no solution.
Although it is quite likely that I've made some mistake along the way, the approach should work.
