Solution:

The answer is $3^{\text{number of 1's in binary expansion of } 2021}=3^{8}=6561$.

Throughout this solution, we say two nonnegative integers overlap in the $2^{\ell}$ position if their binary representations both have a 1 in that position. We say that two nonnegative integers overlap if they overlap in some position. Our central claim is the following.

Claim 1. A triple $(x, y, z)$ is losing if and only if no two of $x, y, z$ overlap.

Let $d_{\ell}(a)$ denote the bit in the $2^{\ell}$ position of the binary representation of $a$. Let $\&$ denote the bitwise and operation: $x \& y$ is the number satisfying $d_{\ell}(x \& y)=d_{\ell}(x) d_{\ell}(y)$ for all $\ell$.

Lemma 1. Let $x, y, z$ be nonnegative integers, at least one pair of which overlaps. Define $x^{\prime}=(x+y) \&(x+z)$ and $y^{\prime}, z^{\prime}$ cyclically. At least one of the inequalities $x<x^{\prime}, y<y^{\prime}, z<z^{\prime}$ holds.

Proof. Let $\ell$ be maximal such that two of $x, y, z$ overlap in the $2^{\ell}$ position. We case on how many of the additions $x+y, x+z, y+z$ involve a carry from the $2^{\ell}$ position to the $2^{\ell+1}$ position, and on the values of $d_{\ell+1}(x), d_{\ell+1}(y), d_{\ell+1}(z)$. Because at least two of $d_{\ell}(x), d_{\ell}(y), d_{\ell}(z)$ equal 1, at least one of the additions $x+y, x+z, y+z$ involves a carry from the $2^{\ell}$ position.

Case 1. One carry.

WLOG let $x+y$ be the addition with the carry. Then, $d_{\ell}(x)=d_{\ell}(y)=1$ and $d_{\ell}(z)=0$. Since $x, y, z$ do not overlap in any position left of the $2^{\ell}$ position, the binary representations of $z, z^{\prime}$ agree left of the $2^{\ell}$ position. As the additions $x+z$ and $y+z$ do not involve a carry from the $2^{\ell}$ position, we have $d_{\ell}(x+z)=d_{\ell}(y+z)=1$, and thus $d_{\ell}\left(z^{\prime}\right)=1$. Thus $z^{\prime}>z$, as desired.

Case 2. At least two carries: $x+y$ and $x+z$ carry and $d_{\ell+1}(y)=d_{\ell+1}(z)=0$, or cyclic equivalent. ($y+z$ may or may not carry.)

Let $i$ be maximal such that $d_{\ell+1}(x)=\cdots=d_{\ell+i}(x)=1$ (possibly $i=0$). By maximality of $\ell, d_{\ell+1}(y)=d_{\ell+1}(z)=\cdots=d_{\ell+i}(y)=d_{\ell+i}(z)=0$. By maximality of $i, d_{\ell+i+1}(x)=0$.

If $d_{\ell+i+1}(y)=d_{\ell+i+1}(z)=0$, then $d_{\ell+i+1}(x+y)=d_{\ell+i+1}(x+z)=1$, so $d_{\ell+i+1}\left(x^{\prime}\right)=1$. The binary representations of $x$ and $x^{\prime}$ agree to the left of the $2^{\ell+i+1}$ position, so $x^{\prime}>x$.

Otherwise, WLOG $d_{\ell+i+1}(y)=1$ and $d_{\ell+i+1}(z)=0$. (Note that, here we in fact have $i \geq 1$.) Then $d_{\ell+i+1}(y+z)=d_{\ell+i+1}(x+z)=1$, so $d_{\ell+i+1}\left(z^{\prime}\right)=1$. The binary representations of $z$ and $z^{\prime}$ agree to the left of the $2^{\ell+i+1}$ position, so $z^{\prime}>z$.

Case 3. At least two carries, and the condition in Case 2 does not occur.

WLOG let $x+y, x+z$ involve carries. Since the condition in Case 2 does not occur, $d_{\ell+1}(y)=1$ or $d_{\ell+1}(z)=1$. In either case, $d_{\ell+1}(x)=0$. WLOG $d_{\ell+1}(y)=1$ and $d_{\ell+1}(z)=0$.

Since the condition in Case 2 does not occur, $y+z$ does not involve a carry from the $2^{\ell}$ position. (Otherwise, $x+y$ and $y+z$ carry and $d_{\ell+1}(x)=d_{\ell+1}(z)=0$.) Then $d_{\ell+1}(x+z)=d_{\ell+1}(y+z)=1$, so $d_{\ell+1}\left(z^{\prime}\right)=1$. The binary representations of $z$ and $z^{\prime}$ agree to the left of the $2^{\ell+1}$ position, so $z^{\prime}>z$.

Proof of Claim 1. Proceed by strong induction on $x+y+z$. There is no base case.

Suppose by induction the claim holds for all $(x, y, z)$ with sum less than $N$. Consider a triple $(x, y, z)$ with $x+y+z=N$.

Suppose no two of $x, y, z$ overlap. If all moves from this position lead to positions with a negative coordinate, $(x, y, z)$ is a losing position, as claimed. Otherwise, the player increases or decreases all coordinates by $k$. Consider the smallest $m$ such that $d_{m}(k)=1$. The player's move will toggle each of $d_{m}(x), d_{m}(y), d_{m}(z)$. Since at most one of the original $d_{m}(x), d_{m}(y), d_{m}(z)$ is 1, at least two of the new $d_{m}(x), d_{m}(y), d_{m}(z)$ will be 1. So, two of the new $x, y, z$ overlap. By induction, the new $(x, y, z)$ is winning. Thus the original $(x, y, z)$ is losing, as claimed.

Conversely, suppose at least one pair of $x, y, z$ overlap. By Lemma 1, at least one of $x<x^{\prime}, y<y^{\prime}, z<z^{\prime}$ holds. WLOG $x<x^{\prime}$. Let the player to move choose $k=x^{\prime}-x$, decrease $y, z$ by $k$, and increase $x$ by $k$. The new coordinates are nonnegative, as
$$y-k=x+y-x^{\prime} \geq 0$$
because $x^{\prime} \leq x+y$, and similarly for the $z$ coordinate. Moreover, the binary representation of the new $x$ consists of the 1's in the binary representations of both $x+y$ and $x+z$; the binary representation of the new $y$ consists of the 1's in that of $x+y$ but not $x+z$; and the binary representation of the new $z$ consists of the 1's in that of $x+z$ but not $x+y$. So, no two of the new $x, y, z$ overlap. By induction, the new $(x, y, z)$ is losing. Thus the original $(x, y, z)$ is winning, as claimed.

We use Claim 1 to count the losing positions $(x, y, z)$ with
$$x+y+z=2021=11111100101_{2}$$
In each position where $d_{\ell}(2021)=0$, losing positions must have $d_{\ell}(x)=d_{\ell}(y)=d_{\ell}(z)=0$. In each position where $d_{\ell}(2021)=1$, the bit triplet $\left(d_{i}(x), d_{i}(y), d_{i}(z)\right)$ is one of $(1,0,0),(0,1,0),(0,0,1)$. This gives a count of $3^{8}=6561$.