  Let $k$ be a positive integer. In this paper, using the modular approach, we
prove that if $k\equiv 0 \pmod{4}$, $30< k<724$ and $2k-1$ is an odd prime
power, then under the GRH, the equation $x^2+(2k-1)^y=k^z$ has only one
positive integer solution $(x,y,z)=(k-1,1,2)$. The above results solve some
difficult cases of Terai's conecture concerning this equation.
