The lhs x^4-6x^2+1 factorises as: (x^2+2x-1)(x^2-2x-1)

Thus (x^2+2x-1)(x^2-2x-1) =7*2^y

Thus, at least one of the factors in the lhs is divisible by 7.

By trial and error, we observe that:

x^2+2x-1=0(mod 7), whenever x=2, or 3

However, x=2 gives the value of the lhs as: -7<0. This is a contradiction.

Thus, x=3, so that:

(3^2+2*3-1)(3^2-2*3-1) = 14*2=28

Thus, 7*2^y = 28,

Or, 2^y=4, giving: y=2

Therefore, (x,y) = (3,2) is the required solution to the given equation.

Note: Haven't been able to prove that x= 2, 3 (mod7), does NOT yield further solutions.

*Edited on ***September 15, 2023, 7:41 am**