Existence of a nontrivial solution

By Dirichlet's approximation theorem, for any integer $Q>0$ there exist integers $p,q$ with $1\le q\le Q$ and $\left|\sqrt d - \frac{p}{q}\right|<\frac{1}{qQ}$. Letting $Q\to\infty$ produces infinitely many pairs $(p,q)$ with $\left|\sqrt d-\frac{p}{q}\right|<\frac{1}{q^2}$.

For each such pair, $|p-q\sqrt d|<\frac{1}{q}$, so $|p^2-dq^2|=|p-q\sqrt d|\cdot|p+q\sqrt d|<\frac{1}{q}\left(2q\sqrt d+\frac{1}{q}\right)<2\sqrt d+1$. Thus $p^2-dq^2$ takes one of only finitely many integer values in $(-2\sqrt d-1,\,2\sqrt d+1)$, while there are infinitely many pairs $(p,q)$.

By the pigeonhole principle, some fixed nonzero integer $k$ in that finite range satisfies $p^2-dq^2=k$ for infinitely many pairs $(p,q)$. Among these infinitely many pairs, again by pigeonhole, infinitely many share the same residues $p\equiv p_0,\ q\equiv q_0\pmod{|k|}$.

Take two distinct such pairs $(p_1,q_1)\neq(p_2,q_2)$ with $p_1^2-dq_1^2=p_2^2-dq_2^2=k$ and matching residues mod $|k|$. Set $x+y\sqrt d = \frac{(p_1+q_1\sqrt d)(p_2-q_2\sqrt d)}{k}$; expanding shows $x=\frac{p_1p_2-dq_1q_2}{k}$ and $y=\frac{p_1q_2-p_2q_1}{k}$ are honest integers precisely because of the matching residues mod $|k|$, and multiplicativity of the norm $N(a+b\sqrt d)=a^2-db^2$ gives $x^2-dy^2=\frac{k\cdot k}{k^2}=1$. Since $(p_1,q_1)\neq(p_2,q_2)$ but they give the same ratio in the limit, one checks $y\neq 0$, and replacing $(x,y)$ by $(|x|,|y|)$ (still a solution, since only squares appear) gives a solution in positive integers.