Show $2^m\le (em)^n \Rightarrow m\leq 2n*\log_2(en)$

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This is a small bonus question in a HW assignment for a machine-learning related course. We need to prove the following for $m,n\geq1$: $$2^m\le (em)^n \Rightarrow m\leq 2n*\log_2(en)$$

A book we use in the course (Understanding Machine Learning: From Theory to Algorithms) mentions that this follows from another lemma:

Let $a\geq 1,b>0$. Then: $$x\geq 4a*\log_2⁡(2a)+2b\Rightarrow x\geq a*\log_2⁡(x)+b$$

However, I was unable to use it since I only get: $$2^m\le (em)^n \Rightarrow m\leq n*\log_2(em)\Rightarrow m\leq n*\log_2(m)+n*\log_2(e)\Rightarrow$$ $$m\leq 4n*\log_2(2n)+2n*\log_2(e)$$

Any hints or directions will be greatly appreciated (we needn't necessarily use the aforementioned lemma, the book might be mistaken here).