By the condition of the problem, we have $a^2 \mid (a^3 + b^3 + c^3)$, $b^2 \mid (a^3 + b^3 + c^3)$ and $c^2 \mid (a^3 + b^3 + c^3)$. Since $a$, $b$ and $c$ are coprime, we see that $a^2 b^2 c^2 \mid (a^3 + b^3 + c^3)$.

Without loss of generality, suppose that $a \ge b \ge c$, so
$$3a^3 \ge a^3 + b^3 + c^3 \ge a^2 b^2 c^2 \Rightarrow a \ge \frac{b^2 c^2}{3},$$
and
$$2b^3 \ge b^3 + c^3 \ge a^2 \Rightarrow 2b^3 \ge \frac{b^4 c^4}{9} \Rightarrow b \le \frac{18}{c^4}.$$
We see that if $c \ge 2 \Rightarrow b \le 1$, the result contradicts $b \ge c$. Thus, $c = 1$.

If $c = 1$ and $b = 1$, then $a = 1$, so $(a, b, c) = (1, 1, 1)$ is a solution.

If $c = 1$, $b \ge 2$ and $a = b$, then $b^2 \mid b^3 + 1$, which is a contradiction!

If $b \ge 2$ and $a > b > c = 1$, then
$$a^2 b^2 \mid (a^3 + b^3 + 1) \Rightarrow 2a^3 \ge a^3 + b^3 + 1 \ge a^2 b^2 \Rightarrow a \ge \frac{b^2}{2},$$
and by $c = 1$,
$$a^2 \mid (b^3 + 1) \Rightarrow b^3 + 1 \ge a^2 \ge \frac{b^4}{4} \Rightarrow 4b^3 + 4 \ge b^4.$$