Using partial fractions, we get
$$
\begin{align}
&\small\frac1{(j+n)^2(k+n)^3}\\
&\small=\color{#00A000}{\frac1{(k-j)^3(j+n)^2}}\color{#C00000}{-\frac3{(j-k)^4(j+n)}}+\frac1{(j-k)^2(k+n)^3}\color{#00A000}{-\frac2{(j-k)^3(k+n)^2}}\color{#C00000}{+\frac3{(j-k)^4(k+n)}}
\end{align}
$$
The red terms cancel each other and the green terms partially cancel each other. Therefore,
$$
\begin{align}
\sum_{j,k,n=1}^\infty\frac1{(j+n)^2}\frac1{(k+n)^3}
&=\sum_{\substack{j,k,n=1\\j\ne k}}^\infty\color{#C00000}{\frac1{(j-k)^2(k+n)^3}}-\color{#00A000}{\frac1{(j-k)^3(k+n)^2}}\\