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Let $k$ be an integer such that $2n^2=dk$. Then we have $$\begin{array}{rl}{k}^2(n^2+d)& =k^2{n}^2+k^{2}d=k^2{n}^2+2k{n}^2\\ & =n^2(k^2+2k)=n^2((k+1{)}^2-1).\end{array}$$ Expression $(k+1{)}^2-1$ is not a square of an integer for $k\in \mathbb{N}$. Then, $k^2(n^2+d)$ is not a square of an integer, so $n^2+d$ cannot be a square of an integer.