Tasty radicals & exponents

Let $\color{#D61F06}{y=2^{\sqrt{x^2-2}+x}}$

From equation

$\begin{aligned}4^{\sqrt{x^2-2}+x}-5\cdot2^{x-1+\sqrt{x^2-2}}=&\ 6\\2^{2(\sqrt{x^2-2}+x)}-5\cdot2^{\sqrt{x^2-2}+x}\cdot2^{-1}=&\ 6\\\left(\color{#D61F06}{2^{\sqrt{x^2-2}+x}}\right)^2-\dfrac52\cdot\color{#D61F06}{2^{\sqrt{x^2-2}+x}}=&\ 6\\\color{#D61F06}y^2-\dfrac52\color{#D61F06}y=&\ 6\\2\color{#D61F06}y^2-5\color{#D61F06}y-12=&\ 0\\(2\color{#D61F06}y+3)(\color{#D61F06}y-4)=&\ 0\\\color{#D61F06}y=&\ -\dfrac32,\color{#D61F06}4\end{aligned}$

But the value of $\color{#D61F06}y$ cannot be a negative. Thus,

$\begin{aligned}2^{\sqrt{x^2-2}+x}=&\ 4\\\color{#624F41}2^{\sqrt{x^2-2}+x}=&\ \color{#624F41}2^2\\\sqrt{x^2-2}+x=&\ 2\\\sqrt{x^2-2}=&\ 2-x\\x^2-2=&\ 4-4x+x^2\\-6=&\ -4x\\x=&\ \dfrac32=\boxed{1.5}\end{aligned}$