$\def\N#1{{\left[\text{#1}\right]}}$tl;dr– There're only 3 possible gates that satisfy these properties:
depend on both of 2 binary inputs to produce 1 binary output;
don't discriminate between the inputs;
aren't just another gate $\texttt{NOT}\text{'d} .$
These 3 gates are usually described as $\texttt{AND} ,$ $\texttt{OR} ,$ and $\texttt{XOR}$ as the $3$ possible gates, along with their $\texttt{NOT}\text{'d} .$ variants, $\texttt{NAND} ,$ $\texttt{NOR} ,$ and $\texttt{NXOR} .$
For each possible input, there's $1$ possible function for each possible output. This means
$$
\N{functions}
~=~
{\N{possible outputs}}
^
{\N{possible inputs}}
\,.
$$
Here, there're two possible outputs, $\left\{0,\,1\right\} ,$ and there're four possible inputs, $\left\{\left\{0,\,0\right\}.\, \left\{0,\,1\right\}, \,\left\{1,\,0\right\}, \, \left\{1,\,1\right\} \right\} ,$ so
$$
\N{functions}
~=~
{2}
^
{4}
~=~
16
\,,
$$
meaning that there're 16 possible gates. as shown @AI0867's answer.
If we exclude both of the inputs, that'd have been $2^1=2$ possible gates. These are the $\texttt{true}$ and $\texttt{false}$ gates, which always return $1$ or $0 ,$ respectively, without regard for their inputs.
If we exclude one of the inputs, that'd have been $2^2 = 4$ possible gates. These would be the identity-gate (return the input), the NOT-gate (invert the input), plus the {true, false}-gates again. So, that's $2$ gates that depend only on one of the inputs.
Since there're 2 inputs, there're:
2 gates that don't care about either input;
2 gates that don't care about the first input; and
2 gates that don't care about the second input;
for a total of 6 gates that don't care about both inputs, reducing the 16 possibilities down to 10 gates which can be meaningfully described as having 2 inputs.
Then, we can note that of these 10 gates, half of them just return the inverted input $(\text{e.g.},$ $\texttt{AND}$ vs. $\texttt{NAND}).$ So we can say that it's just 5 gates, plus their inversions. @AI0867's answer covered this, too.
The point I'd add is that this analysis assumes that inputs are ordered; for example, $\left\{0,1\right\}$ is distinct from $\left\{1,0\right\} .$ However, in practice, we don't tend to discuss gates which discriminate between the inputs.
So, instead of having $4$ distinguishable inputs, we have $3 .$ Then, that's $2^3 = 8$ possible gates. We again remove the $2$ that don't depend either gate; we don't remove the other $2+2=4$ that didn't depend on just one gate because we no longer distinguish between the two gates, so those cases already don't exist. This leaves $2^3-2=6$ possible gates that:
Then, again, we can note the inversion of the outputs, so instead of saying $6$ different gates, we can say that there're $3$ gates with their inversions.
This leaves $\texttt{AND} ,$ $\texttt{OR} ,$ and $\texttt{XOR}$ as the $3$ possible gates, plus their inversions, $\texttt{NAND} ,$ $\texttt{NOR} ,$ and $\texttt{NXOR} .$
In short, when we talk about gates as though only 3 binary-gates exist (plus the unary $\texttt{NOT}$ gate), that's because there're only 3 that:
depend on both inputs;
don't discriminate between the inputs;
aren't just another gate with the output inverted.