Parameterize the solutions in integers to $$x^2 + 5y^2 = z^2$$
To make it easier, consider only the solutions such that the GCD of x, y, and z is 1. Also assume that x, y, and z are positive, and that x is odd. The analysis will probably still give two cases, but we can combine them into one by use of absolute values. Basically I have to prove that the analysis is correct.
Another way to think about such problems is to convert this to $$\left(\frac{a}{c}\right)^2+5\left(\frac{b}{c}\right)^2=1,$$ and then think about rational points (both coordinates rational) on the curve $$x^2+5y^2=1.$$ One obvious point is $(1,0)$. Let us consider a line that passes through this point $y=m(x-1)$ with rational slope $m$. Observe that all the points of intersection of this line and the curve (given above) will be rational. To find the intersection we solve $$x^2+5m^2(x-1)^2=1.$$ This is equivalent to $$x^2(5m^2+1)-10m^2x+(5m^2-1)=0.$$ Since one of the roots of this equation has to be $x=1$ (our trivial rational point) so the other root will be $$\alpha=\frac{5m^2-1}{5m^2+1}.$$ Thus we obtain the following point as the other point of intersection: $$\left(\frac{5m^2-1}{5m^2+1}, \, \frac{-2m}{5m^2+1}\right).$$
So if we take $a=5m^2-1, b=-2m$ and $c=5m^2+1$ (with $m \in \mathbb{Z}$) then we can obtain integer solutions to the equation $a^2+5b^2=c^2$.
More generally, if we take $m=\frac{r}{s}$ (with $r,s \in \mathbb{Z}$ and $s \neq 0$) then we can have $a=5r^2-s^2, b=-2rs$ and $c=5r^2+s^2$ as the solutions.
Suppose that $x^2+5y^2=z^2$ and $(x,y)=1$. Then there are $a,b$ so that $ax+by=1$, and therefore,
$$
\begin{align}
b^2x^2+5b^2y^2&=b^2z^2\tag1\\
b^2x^2+5(1-ax)^2&=b^2z^2\tag2\\
(10a-5a^2x-b^2x)x+(b^2z)z&=5\tag3
\end{align}
$$
Explanation:
$(1)$: multiply $x^2+5y^2=z^2$ by $b^2$
$(2)$: apply $by=1-ax$
$(3)$: rearrange terms
$(3)$ implies that $(x,z)\mid5$. However, if $(x,z)=5$, then $25\mid z^2-x^2=5y^2$, which implies $5\mid y$, but that would mean $5\mid(x,y)$. Therefore, $(x,z)=1$.
Thus, $(z-x,z+x)\mid2(x,z)=2$.
If $(z-x,z+x)=1$, then $$ \overbrace{\ (z\pm x)\ }^{5m^2}\overbrace{\ (z\mp x)\ }^{n^2}=5y^2 $$ $$ (x,y,z)=\tfrac12\left(\pm\left(5m^2-n^2\right),2mn,5m^2+n^2\right) $$ where $(m,n)=1$, $m\equiv n\equiv1\pmod2$, and $n\not\equiv0\pmod5$.
If $(z-x,z+x)=2$, then $$ \overbrace{\ (z\pm x)\ }^{10m^2}\overbrace{\ (z\mp x)\ }^{2n^2}=5y^2 $$ $$ (x,y,z)=\left(\pm\left(5m^2-n^2\right),2mn,5m^2+n^2\right) $$ where $(m,n)=1$, $m+n\equiv1\pmod2$, and $n\not\equiv0\pmod5$.
Putting this all together, we have proven the following
Theorem
All integer triplets, $(x,y,z)$, so that $x^2+5y^2=z^2$, where $(x,y)=1$, are given by: $$ \tfrac{[m+n\text{ is odd}]+1}2\left(\,\left|5m^2-n^2\right|,2mn,5m^2+n^2\,\right) $$ where $(5m,n)=1$ and $[\dots]$ are Iverson Brackets.
Examples $$ \begin{array}{c|c} (m,n)&(x,y,z)\\\hline (1,1)&(2,1,3)\\ (1,2)&(1,4,9)\\ (2,1)&(19,4,21)\\ (1,3)&(2,3,7)\\ (3,1)&(22,3,23)\\ (1,4)&(11,8,21)\\ (2,3)&(11,12,29)\\ (3,2)&(41,12,49)\\ (4,1)&(79,8,81)\\ (5,1)&(62,5,63)\\ (1,6)&(31,12,41)\\ (3,4)&(29,24,61)\\ (4,3)&(71,24,89)\\ (5,2)&(121,20,129)\\ (6,1)&(179,12,181) \end{array} $$
Let $gcd(x,y)=gcd(y,z)=1$
Consider $x^2+5y^2 = z^2$
Divdiding by $y^2$ on both sides for above equation
Let
$$t=\frac{x}{y}$$ & $$u=\frac{z}{y}$$
After Substituting $t,u$ we get
$$t^2+5 = u^2$$
Now we need to solve this Diophantine equation for rational solutions
Put
$$u=\frac{nt+m}{n}$$
After solving for t we get
$$t=\frac{5n^2-m^2}{2mn}=\frac{x}{y}$$ &
$$u=\frac{5n^2+m^2}{2mn}=\frac{z}{y}$$
$x=5n^2-m^2$
$y=2mn$
$z=5n^2+m^2$
Along with the solutions mentioned already, the other primitive solutions are given by $$ x = 2 u^2 + 2uv - 2 v^2 \; , \; y = 2uv+v^2 \; , \; z = 2 u^2 + 2uv + 3 v^2 $$
This comes from Gauss composition; by fairly explicit formulas, the square of any number represented by $2 u^2 + 2uv + 3 v^2$ is then represented by $r^2 + 5 s^2 $
z x y z recipe u v
1 1 0 1 A 1 0
1 1 0 1 RAW
3 -2 1 3 B 0 1
3 -2 1 3 B 0 -1
3 -2 -1 3 B 1 -1
3 2 1 3 RAW
7 2 3 7 B 1 1
7 2 -3 7 B 2 -1
7 2 3 7 RAW
9 -1 4 9 A 2 1
9 -1 -4 9 A 2 -1
9 1 4 9 RAW
21 11 8 21 A 4 1
21 11 -8 21 A 4 -1
21 11 8 21 RAW
21 -19 4 21 A 1 2
21 -19 -4 21 A 1 -2
21 19 4 21 RAW
23 -22 3 23 B 1 -3
23 -22 -3 23 B 2 -3
23 22 3 23 RAW
27 22 7 27 B 3 1
27 22 -7 27 B 4 -1
27 22 7 27 RAW
29 -11 12 29 A 3 2
29 -11 -12 29 A 3 -2
29 11 12 29 RAW
41 31 12 41 A 6 1
41 31 -12 41 A 6 -1
41 31 12 41 RAW
43 38 9 43 B 4 1
43 38 -9 43 B 5 -1
43 38 9 43 RAW
47 2 21 47 B 2 3
47 2 -21 47 B 5 -3
47 2 21 47 RAW
49 -41 12 49 A 2 3
49 -41 -12 49 A 2 -3
49 41 12 49 RAW
61 -29 24 61 A 4 3
61 -29 -24 61 A 4 -3
61 29 24 61 RAW
63 58 11 63 B 5 1
63 58 -11 63 B 6 -1
63 58 11 63 RAW
63 -62 5 63 B 2 -5
63 -62 -5 63 B 3 -5
63 62 5 63 RAW
67 -58 15 67 B 1 -5
67 -58 -15 67 B 4 -5
67 58 15 67 RAW
69 29 28 69 A 7 2
69 29 -28 69 A 7 -2
69 29 28 69 RAW
69 59 16 69 A 8 1
69 59 -16 69 A 8 -1
69 59 16 69 RAW
81 -79 8 81 A 1 4
81 -79 -8 81 A 1 -4
81 79 8 81 RAW
83 38 33 83 B 4 3
83 38 -33 83 B 7 -3
83 38 33 83 RAW
87 -38 35 87 B 1 5
87 -38 -35 87 B 6 -5
87 38 35 87 RAW
87 82 13 87 B 6 1
87 82 -13 87 B 7 -1
87 82 13 87 RAW
89 -71 24 89 A 3 4
89 -71 -24 89 A 3 -4
89 71 24 89 RAW
101 61 36 101 A 9 2
101 61 -36 101 A 9 -2
101 61 36 101 RAW
103 -22 45 103 B 2 5
z x y z recipe u v
