2. Let $(x, y)$ be its positive integer solution, then $x, y$ have the same parity, hence, $\frac{x-y}{2} \cdot \frac{x+y}{2}=$ $5^{2} \times 30^{2 n}=2^{2 n} \times 3^{2 n} \times 5^{2 n+2}$.

We can obtain
$$a_{n}=\frac{1}{2}\left((2 n+1)^{2}(2 n+3)-1\right)=(n+1)\left(4 n^{2}+6 n+1\right)$$
(because $\frac{x-y}{2}<\frac{x+y}{2}$, they are both positive divisors of $2^{2 n} \times 3^{2 n} \times 5^{2 n+2}$.) Note that,
$$\begin{aligned} \left(n+1,4 n^{2}+6 n+1\right) & =(n+1,(4 n+2)(n+1)-1) \\ & =(n+1,-1)=1, \end{aligned}$$

Also, $(2 n+1)^{2}<4 n^{2}+6 n+1<(2 n+2)^{2}$, which means $4 n^{2}+6 n+1$ is not a perfect square. Therefore, $a_{n}$ is not a perfect square.