Rational solutions of $x^2 - 5 = y^2$, $x^2 + 5 = z^2$

Because $n=5$ is a congruent number, there exists a rational $v=x^2$, such that $v\pm 5$ is a positive rational square. All positive rational solutions are in bijection to the rational sides $X,Y,Z$ of a right triangle with area $5$; in general the correspondence is given as follows (for a squarefree congruent number $n$): $$ (X,Y,Z)\mapsto v=(Z/2)^2,\; $$ and conversely $$ v\mapsto X:=\sqrt{v+n}-\sqrt{v-n},\; Y:=\sqrt{v+n}+\sqrt{v-n},\;Z:=2\sqrt{v}. $$ In particular, $n$ is congruent if and only if there exists a rational $v=x^2$ such that $v\pm n$ are squares of rational numbers.

To find all rational solutions, a bijection to certain rational points on the elliptic curve $E_n:y^2=x^3-n^2x$ can be used. The elliptic curve $E_5$ for $n=5$ has rank $1$, and all rational points on it are known. Its Mordell-Weil group is given by $E_5(\mathbb{Q})\equiv \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}$, generated by the points $(0,0)$, $(5,0)$ and $(-4,6)$. Take for example $P=(-4,6)$. Then $2P=((\frac{41}{12})^2,-\frac{62279}{1728})$. Let $x=(\frac{41}{12})^2$. Then $x-5=(\frac{31}{12})^2$ and $x+5=(\frac{49}{12})^2$. This gives your first solution $$ \frac{41}{12},\; \frac{31}{12},\; \frac{49}{12}. $$ We obtain infinitely many others (all in fact) the same way.

Reference (among many others): K. Conrad's article.