Find the number of roots of the equation
$\displaystyle
2^x=x^2+1
$
$\displaystyle 2^x=x^2+1$
$\displaystyle e^{\ln(2^x)} = x^2+1 $
$\displaystyle e^{x\ln(2)} = x^2+1 $
$\displaystyle e^{x\ln(2)} = x^2+1 $
$\displaystyle x = \pm(e^{x\ln(2)} -1) $
Now. The exponential function has only 1 value at any point. There is no x value for which the exponential function has more than 1 value.
So, the RHS, inside the brackets, as x varies, there will only ever be 1 unique solution inside the brackets. However outside the brackets we have written $\displaystyle \pm $ so... can you draw a conclusion from there?
so you only need to show that the equation has at most 3 (real) roots. here's a general fact:
Fact: suppose $\displaystyle a>0$ and let $\displaystyle p(x)$ be a polynomial of degree $\displaystyle n$ with real coefficients. then the equation $\displaystyle a^x=p(x)$ has at most $\displaystyle n+1$ (distinct) real roots.
Proof: if $\displaystyle a=1,$ there's nothing to prove by the fundamental theorem of algebra. for $\displaystyle 0 < a \neq 1$ let $\displaystyle f(x)=a^x - p(x).$ suppose $\displaystyle f(x)=0$ has at least $\displaystyle n+2$ roots. then, by Rolle's theorem, the
equation $\displaystyle f^{(n+1)}(x)=0$ must have at least one real root. but $\displaystyle f^{(n+1)}(x)=a^x(\ln a)^{n+1} \neq 0, \ \forall x \in \mathbb{R}.$ contradiction! $\displaystyle \Box$