I'm reading Qiskit's documentation. https://qiskit.org/textbook/ch-gates/multiple-qubits-entangled-states.html#3.2-Entangled-States- and they show qubits entangled as $|00\rangle$ (both qubits spin down) or $|11\rangle$ (both qubits spin up). Also written as
$$ \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) $$
From my understanding, entanglement requires one particle to be spin up and the other spin down. How is the above entangled state possible?
I think you’re getting confused by the singlet state. This is the state $\frac{1}{\sqrt{2}}(|10\rangle - |01\rangle)$. This is a special state because it has an overall spin quantum number $s=0$, from which other interesting properties arise. For QC, the most interesting properties about this state have to do with expectation values. For an example look at this answer.
However, the singlet state is not the only entangled state. You can define entangled states as those for which you can know everything about the composite system but nothing about the individual parts of it. In other (more mathematical) words, those states that cannot be expressed as product states.
Take a look at the Bell states, they are the maximally entangled states for two qubits. The first Bell state, $|\Phi^+\rangle$, is the one you mention in your question.
Another, very simple, way of think about entangled states is by thinking of states in which the result of measuring one qubit tells you something about the other qubit(s) (see below). This relation doesn’t necessarily need to be an inverse relationship like in the singlet state.
As @Adam Zalcman pointed out in the comments, “the result of measuring one qubit tells you something about the other qubit(s)” can also apply to separable states like $\frac12|00\rangle\langle00|+ \frac12|11\rangle\langle11|$. This last state is a mixed state, not an entangled state. What it is saying is that the only information we have about the system is that it is in the $|00\rangle$ or $|11\rangle$ state, each with $1/2$ probability.
In this case, measuring a $0$ in one qubit will tell you the second qubit is also $0$, and the same for $1$. However, this doesn’t mean the state is entangled. The correlation arises from our ignorance about the system, and it is therefore a classical correlation, not a quantum correlation like with entangled states.
For some more on mixed and entangled states, check out this answer from Physics.SE.
As @epelaaez noted in their answer, by definition, a state is entangled if it cannot be expressed as the product of smaller qubit states. For example, $$ \frac{1}{2}\left(|00\rangle + |01\rangle + |10\rangle + |11\rangle\right) =\left[\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)\right]\left[\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)\right]$$ is not entangled because it factors into the product of $\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)$ with itself.
However, you cannot factor $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ since $$\begin{aligned}\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) &= \frac{1}{\sqrt{2}} (|a_0\rangle + |a_1\rangle)(|b_0\rangle + |b_1\rangle)\\ &= \frac{1}{\sqrt{2}} (|a_0b_0\rangle + |a_0b_1\rangle + |a_1b_0\rangle + |a_1b_1\rangle)\end{aligned}. $$ To get the $|00\rangle$ term, you need $a_0=0$ and either $b_0$ or $b_1$ to equal $0$ (with the other one not present): $$\frac{1}{\sqrt{2}} (|0\rangle + |a_1\rangle)(|0\rangle),$$ but then there is no way to get the $|11\rangle$ term from the factorization.
