= Solution
Since $26\equiv2\pmod4$, the <ring of integers> is $\mathbb Z[\sqrt{26}]$ with <discriminant> $104$. Write $P=(2,\epsilon+1)=(2,\sqrt{26})$ and $Q_\pm=(5,\epsilon\pm1)$. Reduction of $X^2-26$ modulo two gives a <double root>, and modulo five gives the roots $\pm1$. Thus $N(P)=2$, $N(Q_\pm)=5$, and $(5)=Q_+Q_-$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot4=2$. The element $\epsilon+1=6+\sqrt{26}$ belongs to $P$ and $Q_+$ and has <field norm> ten, so equality of <ideal norms> gives
$$
\boxed{(2)=P^2,\quad(5)=Q_+Q_-,\quad(\epsilon+1)=PQ_+.}
$$
The Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every <ideal class> contains an <integral ideal> of <ideal norm> at most five. Three is inert and gives no <prime ideal> of <ideal norm> three; <ideals> of <ideal norm> four are $P^2$, principal. Thus the class group is generated by $P,Q_+,Q_-$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have <field norm> $\pm2$, but $x^2-26y^2=\pm2$ would give $x^2\equiv2$ or $11\pmod{13}$, neither a <quadratic residue>. Hence \b[the <class number> is two].
The unit $\epsilon=5+\sqrt{26}$ has <field norm> $-1$. To show it is fundamental, suppose $1<u<\epsilon$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm1/u$, so $y=(u-u')/(2\sqrt{26})$ is a <positive integer> less than two. Thus $y=1$, and $x^2-26=\pm1$ forces $x=\pm5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of $\epsilon^{-1}$ puts it in $[1,\epsilon)$, proving that all units are $\pm\epsilon^n$.
An element of <field norm> $\pm10$ has <principal ideal> $PQ_+$ or $PQ_-$ by <prime ideal factorization>. The first is generated by $\epsilon+1$; the second by $\epsilon-1=4+\sqrt{26}$, of <field norm> $-10$. Two generators of the same <principal ideal> differ by a unit. Therefore all, and only, the integer solutions are
$$
\boxed{x+\sqrt{26}y=\pm\epsilon^n(\epsilon\pm1),\qquad n\in\mathbb Z.}
$$
The two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two <field norm> signs.
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