A note on the diophantine equation $x^2+b^Y=c^z$

summary:Let $a$, $b$, $c$, $r$ be positive integers such that $a^{2}+b^{2}=c^{r}$, $\min (a,b,c,r)>1$, $\gcd (a,b)=1, a$ is even and $r$ is odd. In this paper we prove that if $b\equiv 3\hspace{4.44443pt}(\@mod \; 4)$ and either $b$ or $c$ is an odd prime power, then the equation $x^{2}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(a,2,r)$ with $\min (y,z)>1$