jmc

We begin by adding 2 to both sides of the equation, $$\begin{array}{rl}23& =x^4+\frac{1}{x^4}\\ \Rightarrow 25& =x^4+\frac{1}{x^4}+2\\ \Rightarrow 25& =x^4+2(x^2)\left(\frac{1}{x^2}\right)+\frac{1}{x^4}\\ \Rightarrow 25& ={\left(x^2+\frac{1}{x^2}\right)}^2\end{array}$$ So $x^2+\frac{1}{x^2}$ is equal to either $5$ or $-5$. Since $x^2+\frac{1}{x^2}$ is the sum of two squares, it cannot be negative. So $x^2+\frac{1}{x^2}=\boxed{5}$.