Flipped

Obviously, $x=2$ is a solution, but also $x=4$ since $2^4 = 4^2$ .

$2^x>x^2$ for $x>4$ ,so there are no other solutions.

For $x<2$ , there is one additional negative solution. $2^0=1>0=0^2$ but then $2^{-1}=\frac 12<1=(-1)^2$ . By the Intermediate Value Theorem , there must be one solution in between and since $x^2$ is strictly decreasing for $x<0$ , this is the only negative solution

Therefore, there are a total of 3 solutions.