Answer: 383

Small Hint:

Define $f(t)=\sqrt{19}+\frac{91}{t}$; the equation says that $f$ applied five times returns $x$

Big Hint:

Use the two fixed points of $f$ and track the ratio $\frac{f(t)-\alpha}{f(t)-\beta}$

Solution:

Let $f(t)=\sqrt{19}+\frac{91}{t},$ and let $\alpha>0>\beta$ be its fixed points. They satisfy $$t^2-\sqrt{19}\,t-91=0.$$ A direct subtraction using $91=\alpha(\alpha-\sqrt{19})=\beta(\beta-\sqrt{19})$ gives $$\frac{f(t)-\alpha}{f(t)-\beta}=\frac{\beta}{\alpha}\,\frac{t-\alpha}{t-\beta}.$$ The given equation is $f^5(x)=x.$ If $x$ were neither $\alpha$ nor $\beta,$ iterating the displayed ratio five times would force $(\frac{\beta}{\alpha})^5=1,$ which is impossible because $\frac{\beta}{\alpha}<0.$ Thus the only roots are $\alpha$ and $\beta.$

Their absolute values sum to $\alpha-\beta,$ the difference of the roots of the quadratic. Hence $$A=\sqrt{(\sqrt{19})^2+4(91)}=\sqrt{383},$$ so $A^2=383.$