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Are there signals for which the Fourier transform is known to exist (perhaps including singularities) and for which the z-transform does not converge?

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There are several signals for which the $\mathcal{Z}$-transform doesn't exist but the (discrete-time) Fourier transform does. One important example is the complex exponential $x[n]=e^{jn\omega_0}$ extending from $-\infty$ to $\infty$. As a consequence, also sinusoidal signals only have a Fourier transform but no $\mathcal{Z}$-transform, as already mentioned in GKH's answer.

Another important group of sequences with only a Fourier transform but no $\mathcal{Z}$-transform are impulse responses of ideal frequency-selective filters, such as ideal low pass, high pass, or band pass filters. The same holds for the impulse responses of the ideal Hilbert transformer and the ideal differentiator.

In general, two-sided signals with a constant envelope (such as the complex exponential and sinusoids), and two-sided signals that decay only as $1/n$ (such as impulse responses of ideal filters) don't have a $\mathcal{Z}$-transform, but they can have a Fourier transform. The corresponding expressions for the Fourier transform may contain Dirac impulses, or they are discontinuous.

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The unit step signal $u[n]$ is the one that comes to my mind. It's DTFT exists with singularity $$ H(e^{j\omega}) = \pi \delta(\omega) + \frac{1}{1-e^{-j\omega}} $$ but Z-transform does not exist because of region of convergence is specified as $|z| \gt 1$ $$ H(z) = \frac{1}{1-z^{-1}},\, |z| \gt 1 $$ So at $|z|=1$, DTFT of $u[n]$ exists with singularity but Z transform does not exists.

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Sinusoidal signals do have a Fourier Transform using Delta distributions but their Z Transform does not converge for any value of $z$.

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