Does Noether theorem apply only to energy conservation?
No.
The way I understand it is that if we imagine a universe where all the laws of physics always require time as one of the inputs, then energy in this universe is not conserved.
When there is explicit time dependence in the system, the energy is generally not conserved. Generally, the Hamiltonian energy will change with time as:
$$
\frac{dH}{dt} = \frac{\partial H}{\partial t}\;,\tag{A}
$$
where the RHS is generally not zero when there is explicit time dependence.
But the laws of physics do not necessarily only describe energy, do they?
That's right, there are other "integrals of the motion," as they say.
There's at least linear/angular momentum, and maybe other things as well.
Yes, linear momentum conservation follows from invariance under spatial translation, just like energy conservation follows from invariance under time translation.
What makes time translation invariance specific to energy?
It just is the way it is. For example, consider a classical real scalar field theory. The conserved charges associated with space-time translations are the $0\mu$ components of the stress-energy tensor, which can be written in this case as:
$$
T^{\mu}_{\nu} = \frac{\partial L}{\partial \partial_\mu\phi}\partial_\nu\phi - L\delta^{\mu}_\nu\;,
$$
from which it follows that
$$
H = \int d^3x T^{00}
$$
is the conserved charge associated with time-translation invariance, i.e., the energy.
And
$$
P^i = \int d^3x T^{0i}
$$
is the conserved charge associated with space-translation invariance, i.e., the momentum.
A simple way to see how time-translation invariance leads to energy conservation is to just use Eq. (A) above.
To put a little more color on that equation, recall that the Hamiltonian energy is defined for a system with a single degree of freedom as
$$
H = p \dot q - L(q,\dot q)\;,\tag{1}
$$
where
$$
p = \frac{\partial L}{\partial \dot q}\;,
$$
is the momentum, and where $q$ is the generalized coordinate.
Note that I wrote $L(q,\dot q)$ in Eq. (1) to indicate that there is no explicit time dependence in the system (so we suspect the energy will be conserved).
From Eq. (1) we have:
$$
\frac{dH}{dt} = \dot p \dot q + p\ddot q - \frac{\partial L}{\partial q}\dot q - \frac{\partial L}{\partial \dot q}\ddot q
$$
$$
= 0\;,\tag{2}
$$
since the second term cancels with the fourth term, by definition, and the first term cancels with the third term, by the Lagrange equations of motion. Eq. (2) says that the energy is conserved. Thus, we see that time translation invariance leads to conservation of energy.
Similarly, suppose the Lagrangian was translationally invariant. Then we have
$$
\frac{\partial L}{\partial q} = 0\;,
$$
which, by the Lagrange equations of motion says:
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot q} = 0 \equiv \frac{dp}{dt}\;.
$$
Thus, we see that space translation invariance leads to conservation of linear momentum.
