By subtracting the equality (i) from (ii) we obtain $(a^2 - c^2) \cdot (b^2 - 2023^2) = 0$. Because $a$, $b$, $c$ are positive integers, we deduce that $a = c$ or $b = 2023$.

If $a = c$, using (i) we find $(a^2 - 2023b)^2 = 0$, that is $a^2 = 2023b = 7 \cdot 17^2 \cdot b$. Consequently $7 \mid b$ and $7 \mid a = c$, therefore the greatest common divisor of $a$, $b$, $c$ and $2023$ is at least $7$, contradicting assumption (iii).

If $b = 2023$, from (i) we obtain $(ac - 2023^2)^2 = 0$, that is $ac = 2023^2 = 7^2 \cdot 17^4$. From (iii) we deduce that $a$ and $c$ are relatively primes, therefore the solutions are: $(a, b, c) \in \{(1, 2023, 2023^2), (7^2, 2023, 17^4), (17^4, 2023, 7^2), (2023^2, 2023, 1)\}$.