Có bao nhiêu số nguyên dương $y$ thỏa mãn bất phương trình $\left[...

Điều kiện: $x>0$.
$\left[ \text{lo}{{\text{g}}_{3}}x-3\text{lo}{{\text{g}}_{\dfrac{1}{3}}}\left( 2x+1 \right)-1 \right]\left( \text{lo}{{\text{g}}_{3}}x-\text{lo}{{\text{g}}_{2}}y+1 \right)<0\Leftrightarrow {{\log }_{3}}\left[ \dfrac{x{{\left( 2x+1 \right)}^{3}}}{3} \right].\left[ {{\log }_{3}}\left( 3x \right)-{{\log }_{2}}y \right]<0$.
Trường hợp 1:
$\begin{aligned}
& \left\{ \begin{aligned}
& {{\log }_{3}}\left[ \dfrac{x{{\left( 2x+1 \right)}^{3}}}{3} \right]<0 \\
& {{\log }_{3}}\left( 3x \right)-{{\log }_{2}}y>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 8{{x}^{4}}+12{{x}^{3}}+6{{x}^{2}}+x-3<0 \\
& 3x>{{3}^{{{\log }_{2}}y}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -1,82<x<0,445 \\
& 3x>{{3}^{{{\log }_{2}}y}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 0<x<0,445 \\
& \dfrac{{{3}^{{{\log }_{2}}y}}}{3}<x \\
\end{aligned} \right. \\
& \Leftrightarrow 1\le y<1,36\Rightarrow y=1\left( y\in {{\mathbb{Z}}^{+}} \right). \\
\end{aligned}$
Trường hợp 2:
$\begin{aligned}
& \left\{ \begin{aligned}
& {{\log }_{3}}\left[ \dfrac{x{{\left( 2x+1 \right)}^{3}}}{3} \right]>0 \\
& {{\log }_{3}}\left( 3x \right)-{{\log }_{2}}y<0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 8{{x}^{4}}+12{{x}^{3}}+6{{x}^{2}}+x-3>0 \\
& 3x<{{3}^{{{\log }_{2}}y}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<-1,82\cup x>0,445 \\
& 3x<{{3}^{{{\log }_{2}}y}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 0,445<x<2022 \\
& 0<x<\dfrac{{{3}^{{{\log }_{2}}y}}}{3} \\
\end{aligned} \right. \\
& \Rightarrow \dfrac{{{3}^{{{\log }_{2}}y}}}{3}\le 2022\Rightarrow 0<y\le 243,7. \\
\end{aligned}$
Kết hợp hai trường hợp có $244$ giá trị $y$ thỏa yêu cầu bài toán.