Assume there is such a divisor and call it, for convenience, $d\in\mathbb{N}^{*}$. Note that since $d^2n^2+d^3=d^2(n^2+d)$ is the square of an integer, then so must be $n^2+d$. Let $n^2+d=m^2$, where $m\in\mathbb{N}^{*}$. Let $2n^2=d\cdot k$. Then $d=\frac{2n^2}{k}$. We now have $n^2+\frac{2n^2}{k}=m^2$. Multiplying each side by $k^2$ we have $n^2\cdot k^2+n^2\cdot 2k=(mk)^2\Leftrightarrow n^2(k^2+2k)=(mk)^2$. Therefore $k^2+2k$ must be a square, but this is impossible since $k^2<k^2+2k<k^2+2k+1=(k+1)^2$ for all $k\in\mathbb{N}^{*}$. Remark. This is very similar to Romania JTST 2019/1. @below Thanks and sorry for my careless mistakes. Fixed.

Your solution is correct except it should be $ n^2 $ $\cdot$ $ k^2$ + $n^2$ $\cdot$ $2k$ $=$ $(mk)^2$ , you had $ (mn)^2$ for RHS. Also, it should be $n^2(k^2+2k)$ $=$ $ (mk)^2$. You had $m^2$ for RHS.

I have a different solution to this problem. Assume for the sake of contradiction that such $(n,d)$ exist. We have $$d^2n^2+d^3=x^2.$$Rewriting it $$n^2+d=\left(\frac{x}{d}\right)^2=r^2,$$where $r\in\mathbb N$. Suppose $d$ has a prime divisor $p$, therefore $p\mid n^2\implies p\mid n$. Also, $p\mid r^2\implies p\mid r$, thus $p^2\mid d$, divide sides by $p^2$ and obtain new one: $$n_1^2+d_1=r_1^2,$$where $n_1=\frac{n}{p},r_1=\frac{r}{p},d_1=\frac{d}{p}$. Repeat the process for all odd divisors of $p$. Thus, after $k$-th step, we have $$n_k^2+d_k=r_k^2,$$for which $d_k$ does not contain any odd primes. Now consider $d=1$, then $1=d=(r-n)(r+n)$, thus $r+n=r-n\implies r=0$, contradiction. Now let us consider case where $\nu_2(d_k)=m$ and since $d_k\mid 2n_k^2$ $\nu_2(n_k^2)\geq m-1$. If $m$ is odd, then $2^{\frac{m-1}{2}}\mid n_k$, therefore, $2^{m-1}\mid r_k^2\implies 2^{\frac{m-1}{2}}\mid r_k$, thus we get $$n_l^2+d_l=r_l^2,$$where $n_l=\frac{n_k}{2^{m-1}},r_l=\frac{r_k}{2^{m-1}},d_l=\frac{d_k}{2^{m-1}}$. Now notice that $d_l$, since $\nu_2(d_k)=m$ and it has no other divisors. If $2\mid n_l$, then we divide this $2$ also out, thus $d_l=2$ and we see that $$2=(r_l-n_l)(r_l+n_l),$$but note that one of the terms must be therefore $1$ and other $2$, which implies that $r_l=\frac{3}{2}$, contradiction. Similarly, if $m$ is even, then $\nu_2(n_k)\geq \lceil{\frac{m-1}{2}}\rceil=\frac{m}{2}$, therefore we have $2^{m}\mid r_k$ and therefore we deduced this also to solving $1=d=(r-n)(r+n)$, which implies $r=0$. We conclude that there is no such $d$, we are done.