Find the number of real values of x satisfying 2^x=x^2+1.

Let $h(x)=2^x-x^2-1$. Clearly $h(0)=0$ and $h(1)=0$, so $x=0,1$ are roots. For $x<0$, $2^x<1\le x^2+1$, so there is no negative root. Also $h(0.5)=\sqrt2-1.25>0$, $h(2)=-1$, $h(3)=-2$, $h(4)=-1$ and $h(5)=6>0$, so $h$ is positive on $(0,1)$, negative on $(1,4]$ and has a further root in $(4,5)$. To see there are no further roots, note $h'''(x)=2^x\ln^32>0$, so $h''$ is strictly increasing and can vanish at most once. Rolle's theorem then bounds the number of zeros of $h$ by $3$ (if $h$ had four zeros then $h'''$ would have a zero). Hence there are exactly three roots: $0$, $1$ and one in $(4,5)$. A numerical scan confirms this.