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For any positive integer $D$ which is not a square, let $(u_1,v_1)$ be the least positive integer solution of the Pell equation $u^2-D{v}^2=1,$ and let $h\left(4D\right)$ denote the class number of binary quadratic primitive forms of discriminant $4D$. If $D$ satisfies $2\nmid D$ and $v_{1}h\left(4D\right)\equiv 0(modD)$, then $D$ is called a singular number. In this paper, we prove that if $(x,y,z)$ is a positive integer solution of the equation $x^y+y^x=z^z$ with $2\mid z$, then maximum $max\{x,y,z\}<480000$ and both $x$, $y$ are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions $(x,y,z)$.