Finding the stress tensor?

There are many exact solutions of the Navier-Stokes equation, which you can find in standard text books. In practice we don't usually specify the velocity field. Instead, we have an equation of state $P(rho,T)$, and a set of boundary and/or initial conditions.

The specific example you provide is very simple (this is known as Poiseulle flow). In order to satisfy no-slip boundary conditions, the top surface has to move at speed $(u,0,0)$. Then your velocity field is a solution to the static continuity and Navier-Stokes equation for a constant pressure. This is the case because the fluid velicity is linear, so the stress tensor, which is build from first derivatives of $vec{u}$, is constant. $$ sigma_{xx}=sigma_{yy}=sigma_{zz}=-P, ; ; ; ; sigma_{xy}=sigma_{yx}=eta u. $$ Then the derivative of the stress tensor (the RHS of the Navier-Stokes equation) is zero. The LHS is zero because the flow is static, $partial_t vec{u}=0$, and the comoving derivative $(vec{u}cdotvecnabla)vec{u}$ also vanishes. As a result the Navier-Stokes equation is indeed satisfied.

I think nobody discuss the stress tensor because it is almost always divided into a pressure field and a diffusion term. Also apart from some simple analytical flow as you expressed or in exercises, I cannot see any practical interest in the determination of the stress tensor. Looking at the pressure field gives all the information, and the stress tensor can be reconstructing later if really needed.

I just looked in the Fluid Mechanics book of Landau-Lifshitz and found an equation using directly the stress tensor for a flow between rotating cylinders (§18 in the 2nd English Edition).