Take the equation mod 4. The left side is congruent to 0 when x is odd and congruent to 1 when x is even. The right side is congruent to 0 when y>=2, congruent to 2 when y=1 and congruent to 3 when y=0.
The only way for both sides to match is when x is odd an y>=2.
Factor x^4-6x^2+1 into (x^2+2x-1)*(x^2-2x-1). Now substitute x=2w+1 and y=z+2 into the factored version of the equation (w and z will be nonnegative integers):
((2w+1)^2+2*(2w+1)-1)*((2w+1)^2+2*(2w+1)-1) = 7*2^(yz+2)
This simplifies to:
(2w^2+4w+1)*(2w^2-1) = 7*2^z
The left side is always odd, thus we must have z=0. Then the two terms of the left are a factorization of 7; but there is only one factorization over positive integers: 7*1.
Then we must have 2w^2+4w+1=7 and 2w^2-1=1. w=1 is the only common root of these equations.
With z=0 and w=1 then x=3 and y=2. The only pair of positive integers (x,y) which satisfy the problem's equation is (3,2).
Another solution
