Solution:
We have the trivial solutions $(a, b, c) = (0, 0, 0)$ and $(a, b, c) = (1, 1, 1)$, as well as the solution $(a, b, c) = (127, 106, 43)$ and its cyclic permutations.

The case $a = b = c = 0$ works. Without loss of generality, $a = \max \{a, b, c\}$. If $b$ and $c$ are both zero, it's obvious that we have no solution. So, via the inequality
$$a^{2} < a^{2} + 2b + c < (a+2)^{2}$$
we find that $a^{2} + 2b + c = (a+1)^{2} \Longrightarrow 2a + 1 = 2b + c$. So,
$$a = b + \frac{c-1}{2}.$$
Let $c = 2k + 1$ with $k \geq 0$; plugging into the given, we find that
$$b^{2} + b + 2 + 5k \quad \text{and} \quad 4k^{2} + 6k + 3b + 1$$
are both perfect squares. Multiplying both these quantities by $4$, and setting $x = 2b + 1$ and $y = 4k + 3$, we find that
$$x^{2} + 5y - 8 \text{ and } y^{2} + 6x - 11$$
are both even squares.

We may assume $x, y \geq 3$. We now have two cases, both of which aren't too bad:

- If $x \geq y$, then $x^{2} < x^{2} + 5y - 8 < (x+3)^{2}$. Since the square is even, $x^{2} + 5y - 8 = (x+1)^{2}$. Then, $x = \frac{5y - 9}{2}$ and we find that $y^{2} + 15y - 38$ is an even square. Since $y^{2} < y^{2} + 15y - 38 < (y+8)^{2}$, there are finitely many cases to check. The solutions are $(x, y) = (3, 3)$ and $(x, y) = (213, 87)$.

- Similarly, if $x \leq y$, then $y^{2} < y^{2} + 6x - 11 < (y+3)^{2}$, so $y^{2} + 6x - 11 = (y+1)^{2}$. Then, $y = 3x - 6$ and we find that $x^{2} + 15x - 38$ is a perfect square. Amusingly, this is the exact same thing (whether this is just a coincidence due to me selecting the equality case to be $x = y$, I'm not sure). Here, the solutions are $(x, y) = (3, 3)$ and $(x, y) = (87, 255)$.

Converting back, we see the solutions are $(0, 0, 0)$, $(1, 1, 1)$ and $(127, 106, 43)$ and its cyclic permutations.