It follows from Bargmann's limit.
The Bargmann bound is well known for a three-dimensional central potential [1, Thm. XIII.9] [2,3]:
$$ N(\ell) < \frac{1}{2\ell + 1} \int_0^\infty r V^-(r) dr $$
Here, $\ell$ is the azimuthal quantum number, $N(\ell)$ the number of bound eigenstates with this quantum number and $V^-(r) = \max\{0,-V(r)\}$.
With a trick, we can apply this to a one-dimensional potential [2, Section III] [1, Problem XIII.22].
The trick is that the 1D Schrödinger equation
$$ (-\partial_x^2 + u) \psi(x) = E\psi(x) $$
is equal to the $\ell=0$ radial Schrödinger equation
$$ (-\partial_r^2 + V(r)) \psi_{rad}(r) = E\psi_{rad}(r) $$
except for the $\psi_{rad}(0) = 0$ boundary condition.
Also, we need to know that the number of negative energy bound states equals the number of nodes of the zero-energy wave function, and the same holds in the 3D case for the number of nodes of the zero-energy radial wave function.
Long story short, we need to count the nodes of the solution to
$$ (-\partial_x^2 + u) \psi(x) = 0 . $$
And we can do so by splitting the problem into two parts that look like counting the negative energy bound states of 3D radial problems.
This is explained in detail in [2, Section III], the result is that
$$ \boxed{N < 1 + \int_{-\infty}^\infty |x| u^-(x) dx } $$
in the 1D-case.
Finally, apply this to your case.
Note that $0 \leq |x| \leq |x|+1$ and $0 \leq u^-(x) \leq |u(x)|$.
Hence,
$$ N < 1 + \int_{-\infty}^\infty (1 + |x|)\, |u(x)|\; dx < \infty , $$
there is a finite number of bound eigenstates.
[1] M. Reed, B. Simon: Methods of Modern Mathematical Physics 4: Analysis of Operators.
[2] K. Chadan, N.N. Khuri, A. Martin, T.T. Wu: Bound States in one and two Spatial Dimensions (arXiv:math-ph/0208011)
[3] https://en.wikipedia.org/wiki/Bargmann%27s_limit