Was posted here, but without proofs https://artofproblemsolving.com/community/q1h2938166p26294612

Suppose the contrary that $n^2+d=(n+k)^2$ for some $k\ge 1$. Then $d=2nk+k^2\mid 2n^2$. Now, let $n=un_1$ and $k=uk_1$ where ${\rm gcd}(n_1,k_1)=1$. We arrive at $k_1(2n_1+k_1) \mid 2n_1^2$. Now, using $k_1\mid 2n_1^2$ and $(k_1,n_1)=1$, it follows that $k_1\mid 2$, so $k_1\in\{1,2\}$. Let $k_1=1$. Then, $2n_1+1\mid 2n_1^2$, which together with $2n_1+1\mid 2n_1^2+n_1$ yields $2n_1+1\mid n_1$, a contradiction. Likewise, if $k_1=2$, then $2n_1+2\mid n_1^2$, and thus $n_1+1\mid n_1^2$, which is again contradictory.