Using Fourier series I have managed to show that
$$ \frac{x^4}{12} = \frac{\pi^2 x^2}{6} + 4 \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4}(1-\cos(nx)) , x \in [-\pi,\pi]$$
From here apparently one need just one step to find an expression for $ \sum_{n = 1}^{\infty} \frac{1}{n^4}$. I need a hint what this step might be.
For $\;x=\pi\;$:
$$\frac{\pi^4}{12}=\frac{\pi^4}6+4\sum_{n=1}^\infty\frac{(-1)^n}{n^4}\left(1-(-1)^n\right)\stackrel{\text{Only odd index matters}}\implies\frac{\pi^4}{96}=\sum_{n=1}^\infty\frac1{(2n-1)^4}$$
and from here:
$$I:=\sum_{n=1}^\infty\frac1{n^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\frac{\pi^4}{96}\implies$$
$$\color{red}{I=\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{96}\cdot\frac{16}{15}=\frac{\pi^4}{90}}$$