Có bao nhiêu giá trị nguyên của tham số $m\in \left( -2022;2022...

Xét $f\left( x \right)={{x}^{3}}+\left( 2m+1 \right)x-2\Rightarrow {f}'\left( x \right)=3{{x}^{2}}+\left( 2m+1 \right)$
TH1: $\left\{ \begin{aligned}
& {f}'\left( x \right)\ge 0,\forall x\in \left( 1;3 \right) \\
& f\left( 1 \right)\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 3{{x}^{2}}+\left( 2m+1 \right)\ge 0,\forall x\in \left( 1;3 \right) \\
& 2m\ge 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \underset{\left[ 1;3 \right]}{\mathop{\min }} {{x}^{2}}\ge -\dfrac{2m+1}{3} \\
& m\ge 0 \\
\end{aligned} \right.$
Vì $\left\{ \begin{aligned}
& m\in \left( -2022;2022 \right) \\
& m\in \mathbb{Z} \\
\end{aligned} \right.\Rightarrow m\in \left\{ 0;1;2;...;2021 \right\}\text{ }\left( 1 \right)$
TH2: $\left\{ \begin{aligned}
& {f}'\left( x \right)\le 0\forall x\in \left( 1;3 \right) \\
& f\left( 1 \right)\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 3{{x}^{2}}+\left( 2m+1 \right)\le 0\forall x\in \left( 1;3 \right) \\
& 2m\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \underset{\left[ 1;3 \right]}{\mathop{\max }} {{x}^{2}}\le -\dfrac{2m+1}{3} \\
& m\le 0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 9\le -\dfrac{2m+1}{3} \\
& m\le 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m\le -14 \\
& m\le 0 \\
\end{aligned} \right.\Rightarrow m\le -14$
Vì $\left\{ \begin{aligned}
& m\in \left( -2022;2022 \right) \\
& m\in \mathbb{Z} \\
\end{aligned} \right.\Rightarrow m\in \left\{ -2021;-2020;...;-14 \right\}\text{ }\left( 2 \right)$
Từ $\left( 1 \right),\left( 2 \right)$ có $4030$ giá trị nguyên $m$ thỏa mãn yêu cầu bài toán.