You might have also meant a related question: why is $\langle m | n\rangle$ equal to zero if $m\ne n$? Like in 3D it's obvious that we just choose three vectors to point in perpendicular directions and their dot products with each other are zero, but can we still do this in this weird quantum space?
Generally these basis states $|m\rangle$ were carefully derived because they have some relationship with a quantum operator that we could call, say, the "unperturbed Hamiltonian operator" $\hat H_0$, such that $\hat H_0 |m\rangle = E_m |m\rangle,$ where $E_m$ is some real energy of the state. Now our quantum algebra demands that $\big(\langle \phi|\chi\rangle\big)^* = \langle \chi|\phi\rangle$ for any vectors $|\chi\rangle$ and $|\psi\rangle$, where the $*$ is complex conjugation $a + i b \mapsto a - i b;$ if there is an operator in the middle this becomes the only slightly more complicated
$$\big(\langle \phi|\hat A|\chi\rangle\big)^* = \langle \chi|\hat A^\dagger |\phi\rangle.$$
I mention this but actually the condition that all the $E_m$ be real goes hand in hand with a property called Hermitian-ness, which states that $\hat H_0 = \hat H_0^\dagger$, so for right now you do not need to think about this conjugate transpose operation $\dagger$ very hard.
Looking at $\langle m | \hat H_0 |n \rangle$ we can therefore see that this must simultaneously be two different numbers:
$$E_n \langle m |n \rangle
= \langle m | \hat H_0 |n \rangle = \big(\langle n | \hat H_0 |m \rangle\big)^* = \big(E_m~\langle n |m \rangle\big)^* = E_m \langle m | n\rangle.$$
That the first of these equals the last of these can also be written as:
$$\big(E_m - E_n\big)~\langle m|n\rangle = 0.$$
Now this is just a multiplication of two complex numbers being zero, and complex multiplication does still have the property that $|z_1~z_2| = |z_1|~|z_2|$ which means that if $z_1~z_2 = 0$ then either $z_1=0$ or $z_2 = 0$ or both.
So this leads to two big possibilities. First, there's the usual case: the energies are different and then $\langle m | n\rangle = 0$. In a very real sense this is the only case we really need to think about, as in physics things tend to be a bit "noisy" and so if $m\ne n$ then there is some noise which stops $E_m$ from being equal to $E_n.$ This insight is not unique to physics; Tadashi Tokieda talks about using it in pure mathematics at length in his topology and geometry lectures that he's put for free online.
But you can be poked mathematically by $E_m = E_n$ while $m \ne n$ and this is called a "degeneracy" in the unperturbed Hamiltonian, and there is a way to deal with it. You see, this is an indicator that you did not pick (we'll say) $|m\rangle$ right when you solved for these functions. In fact these equations are saying that you should have picked $$|m'\rangle \propto |m\rangle - |n\rangle \langle n | m\rangle$$Note that this needs to be renormalized so that $\langle m'|m'\rangle = 1$ but that is no bother. If you tried to do this with $|m\rangle = |n\rangle$ you would get just the zero vector which can't be normalized, but any other choice will lead to a new normalizable vector with $E_{m'} = E_m = E_n$, but with the property that $\langle m'|n\rangle = 0$ as desired.