Since in
$$\hat{H}=\frac{1}{2m} \hat{p}^2 + V(\hat{x}),$$
the first term ("kinetic energy") it's a positive definite operator. For the very artificial potential in question, somewhat sloppily speaking the potential has a minimum at ##x=0## (namely "##-\infty##") there. The energy eigenvalues thus must all be ##>-\infty##, which is hardly a constraint.
To get the eigenvalues you have to solve the time-independent Schrödinger equation,
$$-\frac{\hbar^2}{2m} u_E''(x) -\alpha \delta(x) u_E(x)=E u_E(x).$$
For bound states you need ##E<0##, but for ##E<0## the only solution must be of the form
$$\psi(x)=A \exp(-k|x|), \quad k>0.$$
Now
$$\psi'(x)=-A k \mathrm{sign}(x) \exp(-k |x|)$$
and
$$\psi''(x)=[A k^2 -2 A k \delta(x)] \exp(-k |x|).$$
Plugging this into the Schrödinger equation we get (cancelling the common A exp factor)
$$-\frac{\hbar^2}{2m} [k^2 - 2 k \delta(x)]-\alpha \delta(x)=E.$$
So the only bound-state solution is to make the ##\delta## distribution contribution vanish by choosing
$$\frac{\hbar^2}{m} k = \alpha \;\Rightarrow \; k=\frac{m \alpha}{\hbar^2}.$$
From this we get
$$E=-\frac{\hbar^2}{2m} k^2 = -\frac{m \alpha^2}{2 \hbar^2}.$$
Of course there are plenty of scattering solutions with ##E>0## too.
!