By condition it follows that one of the numbers is divisible by $5$. Moreover, exactly one of the numbers is divisible by $5$, otherwise $(a, b)$ and $90(a, b)$ are divisible by $25$, and so $160$ is divisible by $25$, a contradiction. Let without loss of generality $a \neq 5$, $b \neq 5$. Then $a = 5c$, $(a, b) = (5c, b) = (c, b)$, and the equation can be rewritten as $bc = 32+18(c, b)$. Since $bc$ and $18(c, b)$ are divisible by $(c, b)$, it follows that $(c, b)$ is a factor of $32$, i.e. $(c, b) = 1, 2, 4, 8, 16, 32$.

If $(c, b) \ge 4$, then $bc$ is divisible by $4^2 = 16$, so $18(c, b) = (bc - 32) \neq 16$, and, therefore, $(c, b) \neq 8$. Thus $bc \neq 64$, i.e. $18(c, b) \neq 32$. Therefore, $(c, b) \neq 16$. If either $(c, b) = 16$ or $(c, b) = 32$, then $bc = 16^2$, but $bc = 32+18 \cdot 16$ or $bc = 32+18 \cdot 64$, which are impossible. Hence $(c, b) \le 2$, i.e. is equal to either $1$ or $2$.

1) Let $(c, b) = 1$. Then $bc = 50$, and at least one of the numbers is odd. Since $b \neq 5$, we find that either $c = 25$ and $b = 2$ or $c = 50$ and $b = 1$. This gives the solutions of the initial equation: $(a, b) = (125, 2)$ and $(a, b) = (250, 1)$.

2) Now let $(c, b) = 2$. Then $bc = 32+36=68$. Set $b=2b_1$, $c=2c_1$, where $b_1$ and $c_1$ are coprime. Then $b_1c_1 = 17$. Hence either $c_1 = 1$ and $b_1 = 17$ or $c_1 = 17$ and $b_1 = 1$. This gives the solutions of the initial equation: $(a, b) = (10, 34)$ and $(a, b) = (170, 2)$.

Since the initial equation is symmetric with respect to $a$ and $b$, we have four more solutions: $(34; 10)$, $(2; 125)$, $(2; 170)$, $(1; 250)$.