$x^{2}+2\left( m-1 \right)x+\left( m+5 \right)>0$ $\forall \left( x>1 \right)$

Case-I: $D<0$

$m^{2}-3m-4<0\implies -1\lt m\lt 4$

Case-II: $D\ge 0$

$\implies m\in \left( -\infty ,-1 \right]\cup \left[4,\infty \right)$

$f\left( 1 \right)\ge 0\implies m\ge -\frac{4}{3}$

$-\frac{b}{2a}\lt 1\implies m\gt 0 \implies m\in \left( -1,\infty \right]$