$\because$抛物线$y=\left(x-2\right)^{2}+\left(m+1\right)$，

$\therefore$顶点坐标为$\left(2,m+1\right)$，

$\because$顶点在第一象限，

$\therefore m+1 \gt 0$，

$\therefore m$的取值范围为$m \gt -1$.

故答案为：$m \gt -1$.

$\because$抛物线$y=\left(x-2\right)^{2}+\left(m+1\right)$，

$\therefore$顶点坐标为$\left(2,m+1\right)$，

$\because$顶点在第一象限，

$\therefore m+1 \gt 0$，

$\therefore m$的取值范围为$m \gt -1$.

故答案为：$m \gt -1$.