Is the definition of all non-standard bases dependent on the standard basis? Let $V$ be an $n$-dimensional vector space over field $\mathbb{K}$ and let $\alpha$ be a vector in $V$. Let $B = \{b_1, ..., b_n\}$ be a basis of $V$. When we say $\alpha=(x_1, \ldots, x_n)_B$ we only mean $\alpha= x_1b_1 + \ldots + x_nb_n$. There is nothing else essential to the definition of the coordinates of a vector.
Now, regardles of the basis with respect to which we refer to a vector, such vector is assumed to be the same. Coordinates and basis are a matter of notation, and bear no effect on that which is being denoted.
Now, let us use $\mathbb{R}^2$ to exemplify my doubt. Let us define a non-standard basis for this space; for example, $B = \{(3, 2), (1, 9)\}$. Such definition already depends on the intelligibility of the expressions $(3, 2), (1, 9)$, which are assumed to be coordinates under the standard basis.
Of course one could make the matter uglier by letting $B = \{(x_1, x_2)_{B'}, (y_1, y_2)_{B'} \}$ where $B'$ is another basis of $\mathbb{R}^2$. But the point is the same: the definition of this basis also requires the use of another.
Hopefully my confusion is beginning to be apparent. Yes, formally speaking the definition of a basis is flawless, since it is defined in terms of vectors (not other bases). But when we are faced with actually defining particular bases, we always require another, which raises the question: where did the definition of that other come from?
To conclude, suppose we say $(x_1, x_2)_B = (y_1, y_2)_A$ for basis $A, B$. This can only mean $x_1\textbf{b}_1 + x_2\textbf{b}_2 = y_1\textbf{a}_1 + y_2\textbf{a}_2$. But to actually compute such operation we require $\textbf{b}_i, \textbf{a}_i$ to be defined in terms of some basis as well... Otherwise, how do we know what the scalar multiplications are? So once again, what should that basis be? It is generally assumed to be the standard basis. Which brings the question: is the definition of all non-standard bases dependent on the standard basis?
 A: Just adding this answer to complement the other answers and comments already given:
I think part of the confusion here relates to two different interpretations of tuples. As an element of $\mathbb{R}^2$, the pair $(1,2)$ has first coordinate $1$ and second coordinate $2$ by definition. These coordinates constitute the pair and are intrinsic to it, and do not depend on any basis.
On the other hand, we can also use pairs in $\mathbb{R}^2$ to represent vectors in any $2$-dimensional real vector space $V$ relative to any chosen basis $B$. In this case the coordinates so obtained are not intrinsic to the vectors represented in $V$, because they depend upon the particular basis $B$.
It can get very confusing when we take $V=\mathbb{R}^2$ in the second case here and have multiple possible interpretations for pairs and coordinates within the same space, so it's important to distinguish these different meanings.
A: I think this is the crux of your problem:

But when we are faced with actually defining particular bases, we
always require another,

What you really need when faced with actually defining a particular basis is not another basis, it is some way of actually describing the vectors in the space.
If the space is given to you as $\mathbb{R}^2$ then you are told that each vector is a pair of numbers --- so, as you have observed, essentially its expression in terms of the standard basis.
But suppose the space is the space of all functions that solve the differential equation
$$
y'' + y = 0.
$$
That is a vector space, since the sum of any two solutions is again a solution, as is a scalar multiple of any solution.
There is no "standard basis" to begin with.
It turns out that space is two dimensional, and that
$$
( \sin x ,   \cos x)
$$
is a basis. That basis is in fact the one most commonly used for this space.
A: Vector spaces pre-exist to their basis. E.g. given some set $X,$ the real vector space $\Bbb R^X$ of maps from $X$ to $\Bbb R$ needs no basis for the definition of its operations.
And $\Bbb R^n$ is the particular case $X=\{1,2,\dots,n\}.$
Another particular case, for which there is even no "standard basis", is $X=\Bbb N.$
