= Solution
For a <number field> $K$, let $\mathcal O_K$ be its <ring of integers of a number field>, $r_1$ the number of real embeddings and $r_2$ the number of conjugate pairs of nonreal complex embeddings, so $[K:\mathbb Q]=r_1+2r_2$. The <Dirichlet unit theorem> says
$$
\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r_1+r_2-1},
$$
where $\mu(K)$ is the finite group of <roots of unity> in $K$. Thus an imaginary <quadratic number field> has only roots of unity as units: usually $\{\pm1\}$, with four for $d=-1$ and six for $d=-3$. A real <quadratic number field> has units $\pm\epsilon^n$ for a <fundamental unit> $\epsilon>1$ and integers $n$.
For $K=\mathbb Q(\sqrt{26})$, $\mathcal O_K=\mathbb Z[\sqrt{26}]$ and the <field norm> is $x^2-26y^2$. The element $\epsilon=5+\sqrt{26}$ has norm $-1$ and is a unit. It is the smallest positive unit greater than one. Indeed for $1<u<\epsilon$, its conjugate is $\pm1/u$, so the coefficient $y=(u\mp1/u)/(2\sqrt{26})$ is a positive integer less than two. Therefore $y=1$, and $x^2=26\pm1$ forces $x=5$, giving $u=\epsilon$, a contradiction. The <Dirichlet unit theorem> now gives $\boxed{\mathcal O_K^\times=\{\pm(5+\sqrt{26})^n:n\in\mathbb Z\}}$. Its least positive norm-one generator is $\epsilon^2=51+10\sqrt{26}$.
For the <generalized Pell equation>, write $\alpha=x+y\sqrt{26}$ with norm $N=\pm10$. Change the sign if necessary and multiply by a power of $\epsilon$ to place it in $\sqrt{10}\leq\alpha<\epsilon\sqrt{10}$. These operations preserve integrality and the absolute norm, though the latter operation can change the norm's sign. Since $\alpha'=N/\alpha$, we have $x=(\alpha+N/\alpha)/2\geq0$ and $y=(\alpha-N/\alpha)/(2\sqrt{26})\geq0$. The upper bound gives $y<4$. Checking $y=0,1,2,3$ in $x^2=26y^2\pm10$ leaves only $(x,y)=(4,1),(6,1)$. Both reduced elements lie in the chosen interval. Therefore \b[all integral solutions of either sign are]
$$
\boxed{x+y\sqrt{26}=\pm(4+\sqrt{26})(5+\sqrt{26})^n\quad\text{or}\quad\pm(6+\sqrt{26})(5+\sqrt{26})^n,\qquad n\in\mathbb Z.}
$$
Negative powers are integral because $\epsilon^{-1}=\sqrt{26}-5$. In the first family the norm is $-10(-1)^n$ and in the second it is $10(-1)^n$, so this also separates the two equations. The reduction proves exhaustion rather than merely generating examples.
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