Find the number of real solutions of the equation $2^x+x^2=1$ My attempt is as follows:-
$$2^x+x^2=1$$
$$\left(1+x\cdot log(2)+\frac{x^2\cdot (log(2))^2}{2!}+\frac{x^3\cdot (log(2))^3}{3!}+\dots\right)+x^2=1$$
$$x\cdot log(2)\left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$
$$x=0, \left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$
$$x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}+\dots=-1$$
Now I started to think when this can be negative, it can only be when x is negative.
I also thought about $$S=x^2+\sum_{i=1}^{\infty}\left(\frac{x^i\cdot (log(2))^i}{i+1}\right)$$. But I was not finding the way to transform it into telescopic series.
I am stuck here and not able to proceed further.

Too long for a comment.
Starting from Andrew Chin's analysis, we could obtain a quite good estimate of the solution.
As said, the first derivative cancels at
$$x_*=-\frac{1}{\log (2)}W\left(\frac{\log ^2(2)}{2}\right)$$
Now, develop the function as a Taylor series around this point
$$f(x)=2^x+x^2-1=f(x_*)+\frac 12 f''(x_*)(x-x_*)^2+ O\big((x-x_*)^3\big)$$ Ignoring the higher order terms, the approximation is then
$$x_\pm=x_* \pm \sqrt{-2\frac {f(x_*)}{f''(x_*)}}$$ Computing, this would give as estimates $x_-=-0.5707$ and $x_+=0.0016$ while the exact solutions are $x_-=-0.5722$ and $x_+=0$.
