If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2 = d \cdot k$ and $n^2 + d = m^2$. We substitute to get $n^2 + \frac{3n^2}{k} = m^2$, so $(mk)^2 = n^2(k^2 + 3k)$. We deduce that $k^2 + 3k$ is a perfect square. From the inequalities $k^2 < k^2 + 3k < (k+2)^2$ we deduce that $k^2 + 3k = (k+1)^2$ which implies $k = 1$ and $d = 3n^2$ which verifies the problem.

Alternative Solution:
Let $d$ be a divisor of $3n^2$ such that $n^2 + d = m^2$. We have $m > n$ and $d = (m-n)(m+n)$. Denote $D = (m, n)$, $m = Da$ and $n = Db$. Since $(m-n)(m+n)$ divides $3n^2$, we have $(a-b)(a+b) \mid 3b^2$. The numbers $a$ and $b$ are co-prime, therefore $(a-b)(a+b) \mid 3$. The case $a-b = a+b = 1$ implies $b=0$, which is not possible. The only possibility is $a-b=1$ and $a+b=3$, meaning that $a=2$ and $b=1$. Finally we have $m = 2n$ and $d = (2n)^2 - n^2 = 3n^2$.