I know that in order to make a two quit entangled state, this quantum circuit is used:
But I was wondering if there are any other gate combinations which also create entangled two quit states.
If there are other quantum circuits which achieve this, I was wondering if they would also create any one of the four bell states. Thanks in advance.
Yes, there are many other quantum circuits that prepare entangled states. In fact, in certain sense that can be made precise, almost all possible quantum circuits on two qubits produce entangled states. This corresponds to the fact that almost all pure two-qubit states are entangled. It is the product states that are rare (more precisely: they form a measure zero subset of all pure states).
That said, there are circuits that do not produce any entanglement. For example, any circuit that does not have two-or-more qubit gates or any circuit that consists of single-qubit gates and SWAP gates has this property. Also, running any circuit in reverse destroys the entanglement created by running it forward.
Remark on comparing circuits: Generally speaking, almost any question of the form "Circuit $C$ produces states with property $P$. Are there other circuits that do this?" will have an affirmative answer, because we can always make changes to a circuit by adding or moving around gates without affecting the unitary operation it applies to its input. This is analogous to how we can modify a computer program by adding or moving around instructions without affecting its output. For example, in the circuit in the question we can replace CNOT with CZ sandwiched between two Hadamards on the target qubit or we can add two consecutive X gates anywhere in the circuit.
Example circuit which prepares the Bell state $(|00\rangle + |11\rangle)/\sqrt{2}$ without using the CNOT gate is the following
$$ (S \otimes X) \circ \sqrt{\text{SWAP}} \circ (S \otimes X) $$
where the gates are as described in the Wikipedia article about quantum gates.
The circuit you gave certainly generated an entangled state. In particular, it generates the Bell state: $|\psi\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$. But this is just one particular entangled state, there are many more.
For instance, the following circuit
prepares the state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ which is another Bell state.
Note that if instead of using the Hadamard gate, $H$, I have replaced it with a $RY$ rotation then I would still generate an entangled state.
Since the above circuit will generate a state in the form $|\psi \rangle = \dfrac{ \alpha|01\rangle + \beta|10\rangle}{\sqrt{2}} $ where $|\alpha|^2 + |\beta|^2| = 1$. Also, note that at $\theta = \pi/2$, you get back the first circuit.
You can also use the $RXX$ gate, for example:
This will generate the state $|\psi \rangle = \dfrac{|00\rangle - i|11\rangle}{\sqrt{2}}$ which is also an entangled state. This gate has the matrix representation in the computational basis as $RXX = \begin{pmatrix} \cos(\theta) & 0 & 0 & -i\sin(\theta) \\ 0 & \cos(\theta) & -i\sin(\theta) & 0 \\ 0 & -i\sin(\theta) & \cos(\theta) & 0 \\ -i\sin(\theta) & 0 & 0 & \cos(\theta) \end{pmatrix} $.
