Here's another approach. It depends primarily on the properties of telescoping series, partial fraction expansion, and the following identity for the $m$th harmonic number $\begin{eqnarray*} \sum_{k=1}^\infty \frac{1}{k(k+m)} &=& \frac{1}{m}\sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+m}\right) \\ &=& \frac{1}{m}\sum_{k=1}^m \frac{1}{k} \\ &=& \frac{H_m}{m}, \end{eqnarray*}$ where $m=1,2,\ldots$.
Then,
$\begin{eqnarray*} \sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k} &=& \sum_{k=1}^{\infty} \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n(n+k+2)} \\ &=& \sum_{k=1}^{\infty} \frac{1}{k} \frac{H_{k+2}}{k+2} \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_k +\frac{1}{k+1}+\frac{1}{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\ &=& \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{H_k}{k} - \frac{H_{k+2}}{k+2} \right) + \frac{1}{2} \sum_{k=1}^{\infty} \left(\frac{1}{k(k+1)} + \frac{1}{k(k+2)}\right) \\ &=& \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) + \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) \\ &=& \frac{7}{4}. \end{eqnarray*}$