A while ago, I answered this question here on StackExchange which asks if for any given integer $k$, whether there exists infinitely many natural numbers $n$ such that $$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k) $$

This is indeed the case, and can be shown using the Chinese Remainder Theorem, as in my answer to the linked question. We find infinitely many $n$ such that the common value of $\mu$ evaluated at these points is $0$.

If $k \geq 4$, then one of the values $n+1, n+2, n+3, n+4$ is a multiple of $4$, and hence the common value must, in fact, actually be $0$. For $n<4$, this argument doesn't work, and we can't rule out the possibility that $$ \mu(n+1)=\cdots=\mu(n+k)=\pm 1 $$

My question concerns the case of $k=2$. Are there infinitely many natural numbers $n$ such that either $$\mu(n) = \mu(n+1) = 1$$ or $$\mu(n) = \mu(n+1) = -1?$$