Find explicitly the positive solutions of $2^x=x^2$ 
Find explicitly the positive solutions of the equation $2^x=x^2$

I noticed that $x=2$ and $x=4$ are roots of the equation.
How can I prove that they are the only positive ones? Thanks in advance

They're not the only solutions, if we also allow negative nummbers:. Since $2^0>0^2$ and $2^{-1}<(-1)^2$ there exists a solution $x$ with $-1<x<0$.
Those three are the only real solutions. Let $f(x)=2^x-x^2$. Calculate the third derivative and show that $f'''(x)>0$ for every $x$. If $f$ had four zeroes the Mean Value Theorem would show that $f'$ had three zeroes, hence $f''$ would have two zeroes, hence $f'''$ would have a zero.
So since there are only three real solutions and one is negative there are only two positive solutions.
