On the diophantine equation \(x^ 2+2^ k=y^ n\) (Q1363345)

In a recent paper \textit{J. H. E. Cohn} [Arch. Math. 59, 341-344 (1992; Zbl 0770.11019)] gave a complete solution of the equation \[ x^2+2^k= y^n, \] for \(k\) odd. He proved that if \(k\) is odd there are exactly three families of solutions. In the present paper the authors consider the case \(k\) even and \(m>1\). Their main result is that the equation of the title has two families of solutions given by \[ x=2^m, \qquad y^n=2^{2m+1} \] and by \[ m=3M+1, \quad n=3, \quad x=11\cdot 2^{3M}, \quad y=5\cdot 2^{2M} \] except possibly when \(m=3^{2k+1}m'\), \((m',3)=1\), \(x\) is odd and all prime divisors of \(n\) are congruent to \(7\pmod 8\). The proofs of all results are elementary and rely upon the factorisation of the title equation in the Gaussian field.