Evaluating the infinite sum $ \sum_{k=1}^\infty \frac{H_k}{k^4}$

It seems that your idea is effectively good.

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2k^2(n+k)}=\sum_{n=1}^\infty \frac{1}{n^2}\sum_{k=1}^\infty \frac{1}{k^2(n+k)}$$ $$\sum_{k=1}^\infty \frac{1}{k^2(n+k)}=\sum_{k=1}^\infty\Bigg[-\frac{1}{n^2 k}+\frac{1}{n^2 (n+k)}+\frac{1}{n k^2}\Bigg] =-\frac 1{n^2} \left(\psi (n+1)-\frac{\pi ^2 }{6}n+\gamma \right)$$

Now (using a CAS for the only difficult part), $$\sum_{k=1}^\infty \frac {\psi (n+1) } {n^4}=-\frac{\pi ^2 }{6}\zeta (3)+3 \zeta (5)-\gamma\frac{ \pi ^4}{90}$$

Using a similar idea from this answer - here $$S_4=\sum_{k=1}^\infty \frac{H_k}{k^4}$$

We know,

$$\zeta(n)=\sum_{k=1}^\infty \frac{1}{k^n}\implies \color{blue}{\zeta(n)\zeta(m)=\left(\sum_{k=1}^\infty \frac{1}{k^n}\right)\left(\sum_{k=1}^\infty \frac{1}{k^m}\right)}$$

We can now exploit the above expression,

Consider,

$$2\zeta(2)\zeta(3)=\sum_{x=1}^\infty\sum_{y=1}^\infty\sum_{i=1}^{2}\frac{1}{x^{4-i}y^{i+1}}$$

$$2\zeta(2)\zeta(3)=\sum_{i=1}^{2}\sum_{x=1}^\infty\frac{1}{x^{5}}+2\left(\sum_{y=1}^\infty\sum_{y=x+1}^{\infty}\frac{1}{(y-x)yx^{3}}-\frac{1}{(y-x)xy^{3}}\right)$$

$$2\zeta(2)\zeta(3)=\zeta(5)(2)+2\left(\sum_{y=1}^\infty\sum_{y=x+1}^{\infty}\frac{1}{(y-x)yx^{3}}-\frac{1}{(y-x)xy^{3}}\right)$$

Noting,

$$\sum_{y=1}^\infty\sum_{y=x+1}^{\infty}\frac{1}{(y-x)yx^{3}}=S_4$$

$$\sum_{y=1}^\infty\sum_{y=x+1}^{\infty}\frac{1}{(y-x)xy^{3}}=2S_4-2\zeta(5)$$

$$2\zeta(2)\zeta(3)=\zeta(5)(2)+2\left(S_4-(2S_4-2\zeta(5)\right)$$

$$2\zeta(2)\zeta(3)=\zeta(5)(6)-2S_4$$

$$S_4=\frac12\left({\zeta(5)(6)}-2\zeta(2)\zeta(3)\right)$$

$$S_4=\frac{1}{2}\left({\zeta(5)(6)}-2\zeta(2)\zeta(3)\right)$$

$$\boxed{\color{green}{\sum_{k=1}^\infty \frac{H_k}{k^4}=3\zeta(5)-\zeta(2)\zeta(3)}}$$

Note the following have been used,

Fubini Theorem

Zeta Function

Summation properties

$$\boxed{\color{red}{H_n=\sum_{k=1}^\infty \frac{k}{n(n+k)} }}$$

$$\boxed{\color{red}{\sum_{x=y+1}^{\infty}f(x,y)=\sum_{x=y}^{\infty} f(x)-f(y)}}$$

$$\boxed{\color{red}{\sum_{x=1}^{\infty}\sum_{y=1}^{x-1}=\sum_{x=y+1}^{\infty}\sum_{y=1}^{\infty}}}$$