Xét hàm số  $g\left(x\right)=x^3-3x+m+5$ trên đoạn $\left[0;2\right]$.
Ta có: $g^{\text{'}}\left(x\right)=3x^2-3$.
$g^{\text{'}}\left(x\right)=0\iff \left[\begin{array}{l}x=-1\left(l\right)\\ x=1\left(n\right)\end{array}\right.$.
$g\left(0\right)=m+5;g\left(1\right)=m+3;$$g\left(2\right)=m+7$
Suy ra: $\underset{\left[0;2\right]}{\max }y=\max \left\{\left|m+3\right|;\left|m+7\right|\right\}$.
Trường hợp 1: $\underset{\left[0;2\right]}{\max }y=2\iff \left\{\begin{array}{l}\left|m+7\right|\ge \left|m+3\right|\\ \left|m+7\right|=2\end{array}\right.$$\iff \left\{\begin{array}{l}\left|m+7\right|\ge \left|m+3\right|\\ \left[\begin{array}{l}m=-5\left(n\right)\\ m=-9\left(l\right)\end{array}\right.\end{array}\right.$.
Trường hợp 2:$\underset{\left[0;2\right]}{\max }y=2\iff \left\{\begin{array}{l}\left|m+7\right|\le \left|m+3\right|\\ \left|m+3\right|=2\end{array}\right.$$\iff \left\{\begin{array}{l}\left|m+7\right|\le \left|m+3\right|\\ \left[\begin{array}{l}m=-1\left(l\right)\\ m=-5\left(n\right)\end{array}\right.\end{array}\right.$ .
Vậy $m=-5$ thỏa bài toán.