Let $g\colon \mathbb{R}\to\mathbb{R}$ such that $\log_2(f(x))=g(\log_2(x))$. Then $g(x)=g(x+1)$ and $g(3x)=3g(x)$. I claim for any $x\in \mathbb{Q}^+$, $g(x)=0$. Indeed, let $x=3^a \frac{p}{q}$ where $p,q$ are integers with $\nu_3(q)=\nu_3(p)$. Then $g(x)=3^ag(\frac pq) = 3^a g(\frac pq + \frac{(3^m-1)p}{q}) = 3^{a+m} g(\frac pq)$ where $q\mid 3^m-1$, so $(3^{a+m}-3^a)g(\frac pq)=0$

Let $g:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ be such that $g(x)=f(2^{x})$. Then $g(x)=g(x+1)$ and $g(3a)=g(a)^3$. Then if we pick $k=\frac{3^{\text{ord}_{674}(3)}-1}{674}\in\mathbb{N}$ we have that: \[g\left(\frac{1}{2022}\right)^3=g\left(\frac{1}{674}\right)=g\left(\frac{674k+1}{674}\right)=g\left(\frac{3^{\text{ord}_{674}(3)}}{674}\right)=g\left(\frac{1}{674}\right)^{3^{\text{ord}_{674}(3)}}\]\[\Longrightarrow g\left(\frac{1}{674}\right)=1\Longrightarrow f(\sqrt[2022]{2})=g\left(\frac{1}{2022}\right)=1\]