Answer: $(0,1,0), (-1,0,0)$.

Substituting $x = a - b$, $y = a + b$ we obtain the equation
$$x^3 y^2 = c^2 + 2x + 1,$$
which suffices to be solved in integers such that $x$ and $y$ have equal parity.
We shall consider the equivalent equation
$$x(x^2 y^2 - 2) = c^2 + 1.$$
If both $x$ and $y$ are even then $x^2 y^2 - 2$ is even, whence the l.h.s. is divisible by 4. Thus $c^2 \equiv -1 \pmod 4$, but this is impossible since $-1$ is not a quadratic residue modulo 4.
Let now both $x$ and $y$ be odd; then $x^2 y^2$ is a positive odd number. Consider
two cases:
* If $x^2 y^2 = 1$ then $x^2 = 1$ and $y^2 = 1$. Since $x^2 y^2 - 2 = -1 < 0$ whereas the product $x(x^2 y^2 - 2)$ equals the positive number $c^2 + 1$, we must have $x = -1$. The possibilities $x = -1, y = 1$ and $x = -1, y = -1$ give $a = 0, b = 1$ and $a = -1, b = 0$, respectively. In both cases $c = 0$.
* If $x^2 y^2 > 1$ then $x^2 y^2 - 2 > 0$. Squares of odd numbers give remainder 1 upon division by 4, whence $x^2 y^2 - 2 \equiv 1 \cdot 1 - 2 = -1 \pmod 4$. Consequently there exists a prime divisor $p$ of $x^2 y^2 - 2$ such that $p \equiv -1 \pmod 4$. But then $c^2 \equiv -1 \pmod p$, which is impossible since $-1$ is not a quadratic residue modulo $p$.