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Not an answer, but too long for a comment:

Suppose $n$ is a positive integer such that $\tau(n)=3$ and $\tau(n+1)=6$ and $\tau(n+2)=4$. You already note that then $n=p^2$ for some prime number $p$, and $n+1=2q^2$ for some prime number $q$. It follows that
$$p^2-2q^2=-1,$$
which has the form of a Pell equation. It follows that there is an integer $k$ such that
$$p+q\sqrt{2}=(1+\sqrt{2})(3+2\sqrt{2})^k=(1+\sqrt{2})^{2k+1}.$$
Explicitly the coefficients of the right hand side are given by

\begin{array}{ccc}
x_k &=& \frac{(1+\sqrt{2})^{1-2k}-(1+\sqrt{2})^{2k-1}}{2},\\
y_k &=& \frac{(1+\sqrt{2})^{1-2k}+(1+\sqrt{2})^{2k-1}}{2\sqrt{2}}.
\end{array}

In particular we see that if $2k-1$ is not prime, say $2k-1=ab$ with $a,b>1$, then $x_k$ has two factors
$$(1+\sqrt{2})^a-(1+\sqrt{2})^a
\qquad\text{ and }\qquad
(1+\sqrt{2})^b-(1+\sqrt{2})^b,$$
which are both greater than $2$, and so $x_k$ is not prime, and similarly $y_k$ is not prime. This means there exists a prime number $r$ such that

\begin{array}{ccc}
p &=& \frac{(1+\sqrt{2})^r-(1+\sqrt{2})^{-r}}{2},\\
q &=& \frac{(1+\sqrt{2})^r+(1+\sqrt{2})^{-r}}{2\sqrt{2}}.
\end{array}

A brute force check shows that there is no other solution $n$ with $n\leq10^{200}$.