Note that for $r>0$ one has integral representation
$$J_0(r)=\frac{1}{2\pi}\int_0^{2\pi}e^{ir\cos\phi}d\phi$$
Hence
$$I=\int_0^{\infty}J_0\left(\alpha\sqrt{x^2+z^2}\right)\cos \beta x\,dx=
\frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha\sqrt{x^2+z^2}\cos\phi}\cos\beta x \, d\phi \, dx.\tag{1}$$
On the other hand,
$$\sqrt{x^2+z^2}\cos\phi=z\cos(\phi-\phi_0)+x\sin(\phi-\phi_0),$$
where $\tan\phi_0=-\frac{x}{z}$. Exchanging the order of integration in (1) and shifting $\phi$ by $\phi_0$, we arrive at
$$I=\frac{1}{4\pi}\int_0^{2\pi}\int_{-\infty}^{\infty}e^{i\alpha(z\cos\phi+x\sin\phi)}\cos\beta x \, d\phi \, dx.$$
Finally, using that $\displaystyle\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}dx=\delta(\omega)$ we obtain
$$I=\frac{1}{4}\int_0^{2\pi}e^{i\alpha z\cos\phi}\Bigl[\delta\left(\alpha\sin\phi+\beta\right)+\delta\left(\alpha\sin\phi-\beta\right)\Bigr]d\phi$$
It remains to use $\delta(f(x))=\sum\limits_{\text{zeros of }f}\frac{1}{|f'(x_k)|}\delta(x-x_k)$ and compute the two contributions coming from each of the two delta-functions.