Every positive divisor of $3n^2$, "$d$", can be expressed as $\frac{3n^2}{a}$ where $a \in \mathbb{Z^{+}} $. Then $n^2+d=n^2 \left( \frac{a+3}{a} \right) = x^2$ for some $x \in \mathbb{Z^+} $. Then $\frac{a+3}{a}= \left( \frac{p}{q} \right)^2$ where $ \text{gcd} (p, q) =1$. $$q^2(a+3)=p^2a \Rightarrow q^2|a \; \; p^2|a+3 \Rightarrow a+3=p^2m \;\; a=q^2k \Rightarrow m=k \Rightarrow k(p^2-q^2)=3 \Rightarrow k=1, 3$$Let $k=1$ $$p^2-q^2=(p+q)(p-q)=3 \Rightarrow p=2 \; q=1 \Rightarrow a+3=4a \Rightarrow a=1 \Rightarrow d=3n^2$$Let $k=3$ It is obvious that if $p, q \in \mathbb{Z^+}$ there is no solution to $p^2-q^2=1$. So the only solution is $d=3n^2$ which satisfies the condition.

Trivial with Ukraine TST 2018 p7

Alternatively, set $n^2+d = (n+\ell)^2$ for $\ell\ge 1$, that is $d= \ell (2n+\ell)$. Now, setting $u={\rm gcd}(n,\ell)$ with $n=un_1$ and $\ell=u\ell_1$, we thus have $d=u^2 \ell_1 (2n_1+\ell_1)\mid 3u^2 n_1^2$, yielding $\ell_1 (2n_1+\ell_1)\mid 3n_1^2$. As $(\ell_1,n_1)=1$, we find $\ell_1\in\{1,3\}$. Case 1. $\ell_1=1$. In this case, we find $2n_1+1\mid 3n_1^2$, and since $(n_1,2n_1+1)=1$, we find $2n_1+1\mid 3$. So, $n_1=1$, and therefore $d=3n^2$. Case 2. $\ell_1=3$. In this case, we find $2n_1+3\mid 3n_1^2$. Note that $(n_1,2n_1+3)\in\{1,3\}$ but since $\ell_1=3$ and $(n_1,\ell_1)=1$, we have $3\nmid n_1$, so $2n_1+3$ and $n_1$ are coprime. But then $2n_1+3\mid 3$, not possible.