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 $$x^{3}-3 x^{2}+4=0$$ 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 $\lfloor\sqrt{2021}\rfloor-1=43$.