Solve for $x$: $(x+2)^{\log_2(x+2)}=8(x+2)^2$

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Solve for $x$: $(x+2)^{\log_2(x+2)}=8(x+2)^2$.

Useless attempt:

$$(x+2)^{\log_2(x+2)}=8(x+2)^2,$$
$$(x+2)^{\frac{1}{\log_{x+2}(2)}}=8(x+2)^2.$$
Don't know what to do with this...

I can't figure out a way to solve this equation. I require a hint as to how to proceed with solving it.

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Hint:

Take logarithm with base $2$ on both sides.

we will end up with $$(\log_2 (x+2))^2 = 3 + 2 \log_2 (x+2)$$

now, we just have to solve a quadratic equation.

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Let $\log_2(x+2)=y$, then $x+2=2^y$. The given equation can be written as $$2^{y^2}=8 \cdot 2^{2y} =2^{2y+3}.$$ Now solve $$y^2=2y+3.$$