I assume you want to operate with the given inequality directly.
So, for the induction step $n\to n+1$ you may proceed as follows using the induction hypothesis (IH) $4\log_2(n) \leq n$ for an $n\geq 16$:
$$4\log_2(n+1) = 4 \log_2 \left(n\left(1+\frac 1n\right)\right) = 4 \log_2 n + 4 \log_2 \left(1+\frac 1n\right)$$
$$\stackrel{IH}{\leq} n + 4 \log_2 \left(1+\frac 1n\right)\stackrel{!}{\leq}n+1$$
So, it is enough to show that $4 \log_2 \left(1+\frac 1n\right) \leq 1 \Leftrightarrow \left(1+\frac 1n\right)^4 \leq 2$.
Since $\left(1+\frac 1n\right)^4$ is decreasing, it is enough to show that this inequality is satisfied for $n= 16$, which you can verify easily:
$$\left(1+\frac 1n\right)^4= 1+ \frac{4}{n}+\frac{6}{n^2}+\frac{4}{n^3}+\frac{1}{n^4}$$ $$<1+\frac{4+6+4+1}{n}=1+\frac{15}{n}\stackrel{n = 16}{<}2$$