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$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only one solution $(1,1)$ in $\mathbb N^2$, so $W_\text{fin}=\{1\}$.

Since you explicitly asked for the cardinalities of the respective sets: $|W_0|=0,|W_\text{fin}|=1,|W_\infty|=\aleph_0$.

$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only one solution $(1,1)$ in $\mathbb N^2$, so $W_\text{fin}=\{1\}$.

Since you explicitly asked for the cardinalities of the respective sets: $|W_0|=0,|W_\text{fin}|=1,|W_\infty|=\aleph_0$.

Edit: proof that the pairs described in the first paragraph are, up to symmetry, all the solutions.

Suppose $(k,l),0<k<l$ were a solution not constructed in the above way. Moreover, let this be such a solution with minimal $k$ possible. Then $(n^2k-l,k)$ is a solution. We easily see $0\leq n^2k-l<k$. If $n^2k-l=0$, then we verify that $(k,l)$ was in fact a solution $(n,n^3)$, contradicting choice of $(k,l)$. If it is positive, then by minimality of $k$ it is one of the solutions $(a,b)$ constructed above. But then $(k,l)=(b,n^2b-a)$, again contradicting the choice of $(k,l)$. So the construction gives us all solutions.

$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only one solution $(1,1)$ in $\mathbb N^2$, so $W_\text{fin}=\{1\}$.

$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only one solution $(1,1)$ in $\mathbb N^2$, so $W_\text{fin}=\{1\}$.

Since you explicitly asked for the cardinalities of the respective sets: $|W_0|=0,|W_\text{fin}|=1,|W_\infty|=\aleph_0$.

$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only finitely many solutions, so $W_\text{fin}=\{1\}$.

$W_\infty=\mathbb N\setminus\{1\}$. To see this, note that for $n>1$ we have a solution $(a,b)=(n,n^3)$ with $a<b$, and if we have one such solution, then $(b,n^2b-a)$ is another one (straightforward calculation) with $b<n^2b-a$, from which we easily construct an infinite sequence of distinct solutions.

For $n=1$, we just have to note that $f(a,b)=1$ iff $a^2-ab+b^2=1$ iff $(2a-b)^2+3b^2=4$ which is easily seen to have only one solution $(1,1)$ in $\mathbb N^2$, so $W_\text{fin}=\{1\}$.