How to solve $32^x - 8 = 2 \cdot 4^x$

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I am given the following equation to solve

$$32^x - 8 = 2 \cdot 4^x$$

which one can simplyfy to $$2^{5x}-2^3 = 2^{2x+1}$$

where do we go from here? If we had something like $$2^{2x} - 5 \cdot 2^x + 6 = 0$$

we could convert it to a quadratic, but not in this case.

Any help is highly appreciated.

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You can still convert it into a polynomial, since

$$2^{5x} - 8 = 2\cdot 2^{2x}$$

converts to

$$y^5-8=2y^2$$

if you introduce $y=2^x$.