jmc

We have that $(x^2-5{)}^2=16$ if and only if $x^2-5=\pm 4$. If $x^2-5=4$, then $x^2=9⟹x=\pm 3$, giving 2 solutions. If $x^2-5=-4$, then $x^2=1⟹x=\pm 1$, giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{4}$. Further, the equation is a quartic in $x$, so by the Fundamental Theorem of Algebra, there can be at most four real solutions.