Sketch a B-H hysteresis curve of permanent magnet I have a cylindrical permanent magnet with uniform magnetization $\mathbf{M}=\mathbf{a_z}M$, length $L$ and Diameter $D$.

I also have the following expressions for the magnetic flux density and magnetic field intensity inside the magnet.
$$\mathbf{B}(z)= \frac{\mu_0M}{2} \Bigg[ \frac{z+ \frac{L}{2}}{\sqrt{\big(z+\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}}- \frac{z- \frac{L}{2}}{\sqrt{\big(z-\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}} \Bigg] $$
$$\mathbf{H}(z)=\frac{M}{2} \Bigg[ \frac{z+ \frac{L}{2}}{\sqrt{\big(z+\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}}- \frac{z- \frac{L}{2}}{\sqrt{\big(z-\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}} \Bigg]-M $$
From the current information about the fields can a $\mathbf{B}$-$\mathbf{H}$ hysteresis curve be drawn? And if that's the case how could I draw it? Should I just try to sketch it out with the general knowledge of how a hysteresis could look, or should I try to generate values of the fields and then plot them together?
 A: There's no way from that information to infer the hysteresis curve. Let me explain why.
In general, in a magnetized material, the relationship between magnetic flux density, magnetic field and magnetization is given by (SI units):
$$\boldsymbol{B} = \mu_0(\boldsymbol{H}+\boldsymbol{M})$$
The two relationships that you have written in your question represent exactly that relationship. To draw the hysteresis curve, you would need the additional relationship between $\boldsymbol{M}$ and $\boldsymbol{H}$, and this is missing.
Note also that by assuming a uniform magnetization you get an $\boldsymbol{H}$ field which depends on $z$, and which is actually opposite to the magnetization. This is called the demagnetizing field. Its dependence on $z$ is caused by the boundary conditions that $\boldsymbol{B}$ and $\boldsymbol{H}$ should satisfy at the cylinder's surface. In a real magnet, since $\boldsymbol{M}$ depends on $\boldsymbol{H}$, the magnetization would also depend on $z$. This means that a cylinder cannot support a uniform magnetization. In fact, the ellipsoid is the only known shape that can support a uniform magnetization [1].
So, you can just draw a generic hysteresis curve.
[1] Some sources seem to imply that this is the only shape with this property. The main reference for this seems to be J. A. Osborn , "Demagnetizing Factors of the General Ellipsoid", Phys. Rev., vol. 67, pp. 351-357, 1945. Here the author says (bold mine):

However, for homogeneous bodies whose surface is of second degree,1 $H$ and $J$ (after suitable magnetic treatment)) are uniform throughout [...] The  ellipsoid has the only surface of the second degree that is finite [...]

Reference1 is from Maxwell nonetheless: J. C. Maxwell, Electricity and Magnetism, vol. 2, pp. 66-70, The Clarendon Press, 3rd ed., 1904. Here Maxwell proves that (thanks to hyportnex for digging the result) the (magnetic) potential $V$ must be a quadratic function of the position to have a uniform field inside and then he writes

"Now the only cases with which we are acquainted in which V is a quadratic function of the coordinates within the body are those in which the body is bounded by a complete surface of the second degree, and the only case in which such a body is of finite dimensions is when it is an ellipsoid. We shall therefore apply the method to the case of an ellipsoid.

