$$c^2 + \tau(c^2) - 2 = b^2 \ge (c+1)^2 = c^2 + 2c + 1,$$
which means that $\tau(c^2) > 2c$. All divisors of $c^2$, except $c$, split into pairs of the form $(d, c^2/d)$, where $d < c$ and $c^2/d > c$, so $\tau(c^2) \le 1 + 2(c-1) = 2c - 1$, which contradicts the previously obtained inequality $\tau(c^2) > 2c$. Note also that for $a = 1$ the number $b$ is equal to zero, which is impossible.
Thus $a$ is not a perfect square. Then all positive integer divisors $a$ can be divided into pairs of the form $(d, a/d)$, and hence $\tau(a)$ is an even number. From the equality $a + \tau(a) = b^2 + 2$ we conclude that the numbers $a$ and $b$ have the same parity, so $a + b$ is even.