\(2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 \\ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2(2^{x} -1) \\ 2(2^x- 1) x^2- 2(2^{x} -1) =- (2^{x^2}-2)x \\ 2(2^x- 1) (x^2-1) =- (2^{x^2}-2)x \\\)
By inspection,
If both sides equal zero then they both must be the same.
If x=0 both sides are zero so that is a solution.
If x= +1 or -1 they are solutions too.
If x is bigger than 1 then the left side is positive and the right side is negative so no answers there
If x is less than -1 then the left side is negative and the right side is positive so no answer there either.
So could there be any more answers between -1 and +1?
I do not think so but i do not know how to prove it.
Maybe CuriousDude can show us both.