Diophantine equation from the Latvian Baltic Way team selection competition 2019 So here is the problem statement:
Find all integer triples $(a, b, c)$ such that
$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$
The only things I have so far figured out is that (-1, 0, 0) and (0, 1, 0) are solution, gcd((a-b), c) = 1 and that c must be even.
Any ideas? Maybe even a general solution to the problem?

COMMENT.- (another way) If $a,b,c\in\mathbb Z$ then $a+b,a-b\in\mathbb Z$ so we can study the diophantine equation 
$$d^3s^2=c^2+2d+1$$ Taking now $d$ as a parameter, $x=s$ and $y=c$ we have a family of conics $\Gamma_d$ defined by
$$\Gamma_d:\quad d^3x^2-y^2-(2d+1)=0$$
$d$ cannot be zero clearly in the original equation. 
Whith $d=a-b\gt0$ we have an hyperbola parameterized by
$$x=\sqrt{\dfrac{2d+1}{d^3}}\sec(t)\\y=\sqrt{2d+1}\tan(t)$$
Whith $d=a-b\lt0$ we have an ellipse parameterized by
$$x=\sqrt{\dfrac{2d+1}{d^3}}\cos(t)\\y=\sqrt{2|d|+1}\sin(t)$$
It appears as highly probable that there are not non-trivial solutions.
