I solved the problem (I think) but LaTeX is down and with these Diophantine Equations things get messy.

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,

(x/d)^2+5(y/d)^2=(z/d)^2

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

x^2+5y^2=z^2

Then, gcd(x,y)=gcd(y,z)=1.

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.

Q.E.D.

Lemma 2:In the Diophantine Equation,

x^2+5y^2=z^2

x,y have opposite parity.

ProofIf 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.

Q.E.D.

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

Then,

d|(x+y) and d|[(x+y)-2y]

Thus,

d|(x+y) or d|(2y)

Thus,

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.

Thus, d=2.

Q.E.D.

Hopefully more to come.