Let me speak about it than respond back later.
We note that if gcd(x,y,z)=d then, gcd(x/d,y/d,z/d)=1
Thus, dividing out by d we obtain,
This is the reason why we work with relative primeness because it reduces to find a solution when all are relatively prime.
Here are some useful observations, (which I will use).
Lemma 1: Given the diophantine equations
Proof: Simple, we see that z^2=5y^2+x^2 thus if gcd(x,y)>1 then the right hand side is divisible by it thus z is divisble by it, then gcd(x,y,z)>1 which is not possible.
The other way around we can write,
z^2-5y^2=x^2 thus if gcd(z,y)>1 then the left hand side is divisible by it which means the right hand side it too. Thus, gcd(x,y,z)>1 which is not possible.
Lemma 2: In the Diophantine Equation,
x,y have opposite parity.
Proof If both x,y are even then gcd(x,y)>1 which violate the lemma before . If both are odd then x and y have form of 4k+1 or 4k+3 in any case they lead to a sum of form 4k+2 which is not possible to obtain from squaring a positive integer (z in this case) .
Thus one is even and one is odd.
Note, you ignore the case when y is odd. Yes it is simpler through it. But it can still be odd. Since you asked for that I will base my proof on that fact that y is even.
Lemma 3: If gcd(x,y)=1 and x,y>0 are odd then gcd(x-y,x+y)=2.
Proof: We write, gcd(x+y,[x+y]-2y)=d
d|(x+y) and d|[(x+y)-2y]
d|(x+y) or d|(2y)
d|(x+y) or d|y or d|2
If, d>1 and d|y
Then, d|x which is not possible.
Thus, d does not divide y.
d|2 thus, d=1 or d=2
But d not equal to 1
Because d=gcd(x-y,x+y) and x-y and x+y are both odd.
Hopefully more to come.