We first note that $$10^2 \cdot 30^{2n} = 2^2 \cdot 5^2 \cdot 2^{2n} \cdot 3^{2n} \cdot 5^{2n} = 2^{2n+2} \cdot 3^{2n} \cdot 5^{2n+2}.$$Now we count the number of integer solutions to $x^2-y^2 = 10^2 \cdot 30^{2n}$ by solving the system \begin{align*}x-y &= a\\x+y &= b\end{align*}where $a,b$ are even. This is $2((2n+2)(2n+1)(2n+3)-(2n+1)(2n+3)) = 16n^3+40n^2+28n+6$. Therefore the number of positive integer solutions is $$\dfrac{16n^3+40n^2+28n+6-2}{4} = 4n^3+10n^2+7n+1.$$Now since $4n^3+10n^2+7n+1 = (n+1)(4n^2+6n+1)$ and both $(n+1)$ and $(4n^2+6n+1)$ are relatively prime, it follows that both must be perfect squares. Let $n+1 = m^2$ and then $4n^2+6n+1 = 4(m^2-1)^2+6(m^2-1)+1 = 4m^4-2m^2-1$. We have $$(2x-1)^2 < 4x^2-2x-1 < (2x)^2$$for $x > 1$, so the last expression is not a perfect square.