It is not difficult to see that $\omega(n)\leq 2\log n$, while there exists a constant $A>0$ such that
$$\varphi(n)>A\frac{n}{\log\log n}$$
(see here). Hence,
$$\pi(n)=\varphi(n)-\omega(n)>A\frac{n}{\log\log n}-2\log n$$
but by the PNT $\pi(n)\sim n/\log n$, so for all large $n$, there is an $\epsilon>0$ such that
$$\frac{n}{\log n}-\epsilon<A\frac{n}{\log\log n}-2\log n
\qquad\forall n\geq n_0\text{ satisfying the condition},$$
but this is not possible if you take $n$ large enough. Hence there are only finitely many solutions.