Let's say there is a non-uniform magnetic field $B=(0,0,z)$ (i.e. the field is parallel to $z$-axis and the amplitude is linearly changing along $z$-axis). If there is a current loop (=magnetic dipole) placed in the field with magnetic moment aligned with the field $m=(0,0,1)$, then according to $F=\nabla(m\cdot B)$, the loop shall be experiencing the magnetic force driving it along z axis as we can derive $$F=\nabla(m\cdot B)=\nabla((0,0,1)\cdot (0,0,z))=\nabla z=(0,0,1).$$
On the other hand, every individual moving charge in the loop (let's say it has a shape of a circle) experiences the Lorentz force only in the $xy$-plane (there is no $z$-component of the Lorentz force) and even in the $xy$-plane the Lorentz force has the same magnitude everywhere and the forces applied to the opposite points of the loop cancel out. Thus, the net Lorentz force must be $0$ and the dipole shall not experience any magnetic force.
Best Answer
Your magnetic field $\vec{B} = z \hat{z}$ is not a valid magnetic field, since it has $\vec{\nabla} \cdot \vec{B} = 1 \neq 0$. If you want to have the field have this $z$-dependence, you must also have a non-trivial field in the $xy$-plane. And it's this latter component of the field that causes the force.
For example, if the field is radially symmetric about the center of the loop, you could have $$ \vec{B} = z \hat{z} - \frac{1}{2} r \hat{r} $$ where $r$ is the cylindrical coordinate (distance from the $z$-axis). This field does obey $\vec{\nabla} \cdot \vec{B} = 0$; and the force on the loop due to this new field is $$ \vec{F} = I \oint d\vec{l} \times \vec{B} = I \int_0^{2\pi} (r d \theta \, \hat{\theta}) \times \left(- \frac{1}{2} r \, \hat{r} \right) = \frac{1}{2} I r^2 \hat{z} \int_0^{2\pi} d \theta = I \pi r^2 \hat{z} = I A \hat{z} $$ as expected.
More generally, the Lorentz force law says that the force on any current configuration is $$ \vec{F} = \int \vec{J} \times \vec{B} \, d \tau. $$ For a "small" current configuration it is possible to show that this is equal to the magnetic dipole force formula $\vec{\nabla} (\vec{m}\cdot\vec{B})$; but one step in the derivation relies on the fact that $\vec{\nabla} \cdot \vec{B} = 0$ (as well as $\vec{\nabla} \cdot \vec{J} = 0$.) I believe this derivation is in Zangwill's Modern Electrodynamics if you want to see it, but my recollection is that it's pretty hairy algebraically.