let a>=b ,and
a^2 + 6b^2 = m^2
b^2 + 6a^2 = n^2 , for some integers m and n.
also,
=>7(b^2 + a^2) = m^2 + n^2
-->b^2 + a^2 = (m^2 /7) + (n^2 /7)
thus both m^2 and n^2 must be divisible by 7 for LHS to be an integer.
let m=7p and n=7q
=> b^2 + a^2 = 7(p^2 + q^2)
again the same thing i.e. b^2 and a^2 should be divisible by 7.
this process will go on forever,
I am unable to highlight the conclusion as to why (0,0) is the only solution, though logically, its quite evident that it is the only soln.