Solution:

Let the middle number be $n$, so the numbers are $n-1$, $n$ and $n+1$. Since $4$ does not divide any of them, $n \equiv 2 \pmod{4}$. Furthermore, neither $3$, $5$ nor $7$ divides $n$. Also $n+1+11 \equiv 2 \pmod{4}$ cannot be a square. Then $3$ must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \mid n+1$ which implies $3 \mid n+10$ (a square), so $9 \mid n+10$ hence $9 \mid n+1$, which is impossible. Thus must be $n+11 = m^2$.

Further, $7$ does not divide $n-1$, nor $n+1$, because $1+11 \equiv 5 \pmod{7}$ and $-1+11 \equiv 3 \pmod{7}$ are quadratic nonresidues modulo $7$. This implies $5 \mid n-1$ or $5 \mid n+1$. Again, since $n+11$ is a square, it is impossible $5 \mid n-1$, hence $5 \mid n+1$ which implies $3 \mid n-1$. This yields $n \equiv 4 \pmod{10}$ hence $S(n+1) = S(n) + 1 = S(n-1) + 2$ ($S(n)$ is sum of the digits of $n$).

Since the three numbers are square-free, their numbers of positive divisors are powers of $2$. Thus, we have two even sums of digits - they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3 \mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1) = 2^p$. Consequently $S(n) = 2^p - 1$ is a prime, so $p$ is a prime number. One easily verifies $p \neq 2$, so $p$ is odd implying $3 \mid 2^p - 2$. Then

$\sigma\left(2^p - 2\right) \geq \left(2^p - 2\right)\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6}\right) = 2\left(2^p - 2\right)$. Since this number is perfect, $\frac{2^p - 2}{6}$ must be one, i.e. $p = 3$ and $S(n-1) = 6$, $S(n) = 7$ and $S(n+1) = 8$.

Since $4$ does not divide $n$, the $2$-digit ending of $n$ must be $14$ or $34$. But $n = 34$ is impossible, since $n+11 = 45$ is not a square. Hence, $n = 10^a + 10^b + 14$ with $a \geq b \geq 2$. If $a \neq b$, then $n$ has three digits $1$ in its decimal representation, which is impossible. Therefore $a = b$, and $n = 2 \cdot 10^a + 14$. Now, $2 \cdot 10^a + 25 = m^2$, hence $5 \mid m$, say $m = 5t$, and $(t-1)(t+1) = 2^{a+1} 5^{a-2}$. Because $\gcd(t-1, t+1) = 2$ there are three possibilities:

1) $t-1 = 2$, $t+1 = 2^a 5^{a-2}$, which implies $a = 2$, $t = 3$;
2) $t-1 = 2^a$, $t+1 = 2 \cdot 5^{a-2}$, so $2^a + 2 = 2 \cdot 5^{a-2}$, which implies $a = 3$, $t = 9$;
3) $t-1 = 2 \cdot 5^{a-2}$, $t+1 = 2^a$, so $2 \cdot 5^{a-2} + 2 = 2^a$, which implies $a = 2$, $t = 3$, same as case 1.

From the only two possibilities $(n-1, n, n+1) = (213, 214, 215)$ and $(n-1, n, n+1) = (2013, 2014, 2015)$ the first one is not possible, because $S(215) = 8$ and $\tau(215) = 4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: $2013$, $2014$, $2015$.