For which values of $a$ and $b$, integers $\geq 2, ab-1$ is divisible by the product $(a-1)(b-1)$?
What integer values can $q=\frac{a b-1}{(a-1)(b-1)}$ take? On one hand, $a b-1 > (a-1)(b-1) = a b - a - b + 1$ because $a + b > 2$, so $q > 1$. On the other hand, $\frac{a b-1}{(a-1)(b-1)} < \left(\frac{a}{a-1}\right) \cdot \left(\frac{b}{b-1}\right) \leq 4$ because for any integer $a \geq 2, \frac{a}{a-1} = 1 + \frac{1}{a-1} \leq 2$. In the interval $1 < q < 4$, there are only two possible integers: 2 and 3.

The value 2 is achieved when $a b - 1 = 2(a-1)(b-1)$, which simplifies to $a b - 2a - 2b + 3 = 0$, or $a = \frac{2b-3}{b-2}$. For $\frac{2b-3}{b-2}$ to be an integer, $2b-3$ must be divisible by $b-2$, and thus $2b-3 - 2(b-2) = 1$ must also be divisible by $b-2$. Since 1 has only the divisors 1 and -1, and $b-2 = -1$ has no solution $\geq 2$, the only possible value for $q=2$ is $b=3$, so $a=3$.

The value 3 is achieved when $a b - 1 = 3(a-1)(b-1)$, which simplifies to $a = \frac{3b-4}{2b-3}$. If $2b-3$ divides $3b-4$, it also divides $1 = 2(3b-4) - 3(2b-3)$, and the only solution is $b=2$, so $a=2$.

The only two pairs of solutions are therefore $(2,2)$ and $(3,3)$.