Solution:

Let $n^{2} a-1 = (n x_{n} + 1) d_{n}$, where $x_{n}, d_{n} \in \mathbb{N}$. Then $d_{n} \equiv -1 \pmod{n}$, so
$$n^{2} a-1 = (n x_{n} + 1)(n y_{n} - 1) \quad \text{for some } x_{n}, y_{n} \in \mathbb{N}$$
which reduces to $n a - n x_{n} y_{n} = y_{n} - x_{n} > -x_{n} y_{n}$. From this we obtain $x_{n} \leqslant x_{n} y_{n} < \frac{n}{n-1} a \leqslant 2a$. It follows that in the sequence $x_{1}, x_{2}, \ldots$ there exists a term that occurs infinitely many times. Let us denote that term by $X$. Then $n X + 1 \mid n^{2} a - 1$ and hence
$$n X + 1 \mid X^{2}(n^{2} a - 1) - a(n^{2} x^{2} - 1) = a - X^{2}$$
for infinitely many numbers $n$. This is possible only for $a - X^{2} = 0$, i.e. $X^{2} = a$.

Second solution. As in the first solution, let $n^{2} a-1 = (n x_{n} + 1)(n y_{n} - 1)$, i.e. $y_{n} - x_{n} = n(a - x_{n} y_{n}) = n d_{n}$. We distinguish three cases.
(1) If $d_{n} > 0$, then $a = d_{n} + x_{n}(x_{n} + n d_{n}) > n d_{n} x_{n}$, which is impossible for $n \geqslant a$.
(2) If $d_{n} < 0$, then $a = d_{n} + y_{n}(y_{n} - n d_{n}) = y_{n}^{2} - d_{n}(n y_{n} - 1) > n y_{n} - 1$, which is impossible for $n \geqslant a + 1$.
(3) If $d_{n} = 0$, then $a = x_{n}^{2}$, a perfect square.