$\psi_\alpha$, $\alpha=1,2,\ldots d$, is a column vector of a $d\times d$ unitary matrix $U$; averaging over the Haar measure gives
$$\int d\psi\, \psi_{\alpha} \psi^\ast_\beta \psi_{\alpha'}\psi^\ast_{\beta'}=\frac{1}{d+d^2}\left(\delta_{\alpha\beta}\delta_{\alpha'\beta'}+\delta_{\alpha\beta'}\delta_{\alpha'\beta}\right)$$

so your integral equals

$$\int \mathrm{Tr}( \psi \psi^* \, \, w x^* \,\, \psi \psi^* \,\, y z^*) d\psi=
\langle z|\psi\rangle\langle \psi|w\rangle\langle x|\psi\rangle\langle \psi|y\rangle$$
$$\qquad\qquad = \frac{1}{d+d^2}\left(\langle z|w\rangle\langle x|y\rangle+ \langle z|y\rangle\langle x|w\rangle\right)$$