Belarusian Mathematical Olympiad

Let $[a, b] = x$, and $(a, b) = y$. It is known (and easy to prove) that $ab = [a, b] \cdot (a, b)$, so $ab = xy$. Then the initial equation can be rewritten in the form
$$xy = 300 + 7x + 5y \Leftrightarrow xy - 7x - 5y + 35 = 335 \Leftrightarrow x(y - 7) - 5(y - 7) = 335 \Leftrightarrow (x - 5)(y - 7) = 5 \cdot 67.$$
Factors on the left-hand side are nonnegative integers, and since $x = [a, b] \ge (a, b) = y$, we have $(x - 5) > (y - 7)$. So we have two possibilities.

1) $x - 5 = 67$ and $y - 7 = 5$. Then $x = 72$ and $y = 12$. It means that $a = 12n$, $b = 12m$ for some relatively prime $n, m$, and $ab = xy = 72 \cdot 12$, i.e. $12n \cdot 12m = 72 \cdot 12$, thus $nm = 6$. Taking into account that $a \le b$ and so $n \le m$, we obtain $n = 1, m = 6$. Therefore $a = 12, b = 12 \cdot 6 = 72$, or $n = 2, m = 3$, so $a = 12 \cdot 2 = 24$, $b = 12 \cdot 3 = 36$.

2) $x - 5 = 335$ and $y - 7 = 1$. Then $x = 340$ and $y = 8$. But it is impossible since $[a, b] = x$ must be divisible by $y = (a, b)$, but $340 \nmid 8$.

Therefore, we have two required pairs $(12, 72)$, $(24, 36)$.