A new conjecture concerning the Diophantine equation $x^2 + b^y = c^z$

Abstract

In this paper, using a recent result of Bilu, Hanrot and Voutier on primitive divisors, we prove that if $a = |V_r|$, $b = |U_r|$, $c = m^2 + 1$, and $b \equiv 3 \pmod{4}$ is a prime power, then the Diophantine equation $x^2 + b^y = c^z$ has only the positive integer solution $(x,y,z) = (a,2,r)$, where $r > 1$ is an odd integer, $m \in \mathbf{N}$ with $2 \mid m$ and the integers $U_r$, $V_r$ satisfy $(m + \sqrt{-1} )^r = V_r + U_r \sqrt{-1}$.