Proof that Dirichlet series $\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^2}=\frac{5}{2}$

So I want to prove the following:

$$\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^2}=\frac{5}{2},$$ where $\omega(n)$ is the number of distinct prime factors of $n.$

I computed it to $10^{10}$ and it does seem to be slowly approaching $\frac{5}{2}.$ Also, I am aware of the following result:

$$\sum_{n=1}^{\infty}\frac{\omega(n)}{n^2}=\zeta(2)P(2),$$ where $P(2)$ is the prime zeta function.

I am not quite sure how exactly to go about this, there doesn't seem to be a way to get from the known result to the one I'm trying to solve.

$2^{\omega}=1*\mu^2$ therefore $$ \sum_{n=1}^{+\infty}\frac{2^{\omega(n)}}{n^2}=\sum_{n=1}^{+\infty}\frac{1}{n^2}\sum_{n=1}^{+\infty}\frac{\mu(n)^2}{n^2}=\zeta(2)\frac{\zeta(2)}{\zeta(4)}=\frac{\pi^2}{6}\frac{\pi^2/6}{\pi^4/90}=\frac{5}{2} ​$$ Indeed, $$ \zeta(2s)\sum_{n=1}^{+\infty}\frac{\mu(n)^2}{n^s}=\sum_{n,m=1}^{+\infty}\frac{\mu(n)^2}{(nm^2)^s}=\sum_{p=1}^{+\infty}\frac{1}{p^s}=\zeta(s) $$ because every integer $p\geqslant 1$ can be uniquely written as $p=nm^2$ where $m\geqslant 1$ and $n\geqslant 1$ is a squarefree integer.