The definition that you are using is not the most general. If you insist on applying the way you do, it only applies for a uniform flow in between two counter-mooving walls. Then, the velocity profile is indeed linear.
A Newtonian fluid is defined by the approximation that local stress (or drag) is proportional to local strain. I would write your equation as,
$$ \tau_{ij}(\vec{y}) = \mu \left(\frac{\partial u_i}{\partial y_j}(\vec{y}) + \frac{\partial u_j}{\partial y_i}(\vec{y}) \right)\, . \hspace{2cm} (*)$$
$\vec{u}(\vec{y})$ is the local velocity field. Note that $\tau_{ij}(\vec{y})$ can be a complicated function of $\vec{y}$. The left hand side of equation $(*)$ is the stress on the fluid at position $\vec{y}$. It has two indices because strain affects each component of $\vec{u}(\vec{y})$ in each direction differently. Under stress alone the local velocity field will change according to,
$$ \partial_t u_i(\vec{y}) = \sum_j \partial_{y_j} \tau_{ij}(\vec{y}) \, . \hspace{2cm} (**)$$
The general case will of course involve the full Navier-Stokes equation with this term included. Note that $(**)$ is just a definition of $\tau_{ij}(\vec{y})$ which applies in the most general cases. It just states that the local velocity field will be affected by its neighbours.
The right-hand side of $(*)$ represents the the Newtonian approximation. It is natural to assume that if the fluid is uniform (i.e. its spatial derivatives vanish) it will not be under strain. If it is not uniform (i.e. when different parts of the fluid move at different speeds) however there will be a strain and $\tau_{ij}(\vec{y})$ may be a complicated function of the derivatives of $\vec{u}(\vec{y})$. Assuming that this function can be Taylor expanded, the leading term (for small inhomogeneities, $\partial_{y_i}u_j(\vec{y})$ small) is given by the right hand side of equation $(*)$.
Note that your equation follows from $(*)$ if the fluid only changes in one direction, $\vec{u}(x,y,z) = \vec{u}(x)$.
