Consider this equation as a square equation with respect to $c$. Then,
$$c^2 - abc + (a + b) = 0 \Rightarrow D = (ab)^2 - 4(a + b) = m^2.$$
If $b = 1$, $D_1 = a^2 - 4a - 4 = (a - 2)^2 - 8 = x^2 - 8$, where $x = a - 2$.
If $x = 3$, $D_1 = 1$ is a perfect square, so $a = 5$, $c^2 - 5c + 6 = 0$, which yields the answers: $(5, 1, 2)$ and $(5, 1, 3)$, and corresponding symmetric tuples: $(1, 5, 2)$ and $(1, 5, 3)$.
If $x = 4$, $D_1 = 8$ is not a perfect square.
If $x \ge 5$, $x^2 > D_1 > (x - 1)^2 = x^2 - 2x + 1$, so $D_1$ is not a perfect square.

Now, suppose $b \ge 2$. Then, $2ab \ge a + b + 4 \Leftrightarrow a(b - 1) + b(a - 1) \ge 4$. Hence,
$$(ab)^2 > D = (ab)^2 - 4(a + b) \ge (ab - 4)^2 = (ab)^2 - 8ab + 16.$$
Thus, 4 cases remain.
$D = (ab - 4)^2$, then, from previous estimates, this is only possible if $a = b = 2$, thus, $c = 2$, and we have the solution $(2, 2, 2)$.
$D = (ab - 3)^2$ and $D = (ab - 1)^2$ are impossible, since then $D = (ab - n)^2 = (ab)^2 - 4(a + b) \Leftrightarrow 2abn = n^2 + 4(a + b)$, which gives a contradiction for an odd $n$.

$D = (ab - 2)^2$, then $4ab = 4 + 4(a + b) \Rightarrow ab - a - b - 1 = 0 \Rightarrow a(b - 1) - (b - 1) = 2 \Rightarrow (a - 1)(b - 1) = 2 \Rightarrow a - 1 = 2$ and $b - 1 = 1$. Thus, $a = 3$ and $b = 2$, and we find $c$

from the equation: $c^2 - 6c + 5 = 0$. This gives us the solutions $(3, 2, 1)$ and $(3, 2, 5)$, and their symmetric counterparts $(2, 3, 1)$ and $(2, 3, 5)$.