Too Big to Solve!

$\sqrt{x+\sqrt{4x+\sqrt{16x+\sqrt{...+\sqrt{4^{2019} x+3}}}}}=1+\sqrt{x}$

First Square both sides

$x+{\sqrt{4x+\sqrt{16x+\sqrt{...+\sqrt{4^{2019} x+3}}}}}=1+2\sqrt{x}+x$

${\sqrt{4x+\sqrt{...+\sqrt{4^{2019} x+3}}}}=1+2\sqrt{x}$

Now Square both sides again

$4x+\sqrt{16x+\sqrt{...+\sqrt{4^{2019} x+3}}}=1+4\sqrt{x}+4x$

$\sqrt{16x+\sqrt{...+\sqrt{4^{2019} x+3}}}=1+4\sqrt{x}$

We can see a pattern here, so generalizing it as

$1+2^{n}\sqrt{x}=\sqrt{4^{n} x+\sqrt{...+\sqrt{4^{2019} x+3}}}$

Therefore

$1+2^{2019} \sqrt{x} =\sqrt{4^{2019} x+3}$

square both sides,

$1+2\times2^{2019}\sqrt{x }+\bigg(2^{2019}\bigg)^{2} x =4^{2019}x +3$

$1+2^{2020}\sqrt{x} +2^{4038} x =4^{2019}x +3$

But, $4^{2019}=2^{4038}$

$1+2^{2020}\sqrt{x} =3$

$2^{2020}\sqrt{x} =2$

$\sqrt{x } =\dfrac{2}{2^{2020}}$

$\sqrt{x} =\dfrac{1}{2^{2019}}$

$x =\dfrac{1}{2^{4038}}$