, 2021

Solution:

Taking $\log_{a}$ of both sides and simplifying gives
$$-4 \log_{b} a = \left(\log_{a} b\right)^2 - 3 \log_{a} b$$
Plugging in $x = \log_{a} b$ and using $\log_{b} a = \frac{1}{\log_{a} b}$ gives
$$x^3 - 3x^2 + 4 = 0$$
We can factor the polynomial as $(x-2)(x-2)(x+1)$, meaning $b = a^2$ or $b = a^{-1}$. The second case is impossible since both $a$ and $b$ are positive integers. So, we need only count the number of $1 < a, b \leq 2021$ for which $b = a^2$, which is $\lfloor \sqrt{2021} \rfloor - 1 = 43$.