We summarize why the Hilbert scheme of points \(\mathrm{Hilb}^n(\mathbb{C}^2)\) is a quiver variety here.
The Quiver
Consider the following quiver representation with dimension vector \(v=(1,n)\), where the left vertex is a framing vertex and only the right vertex receives group actions.
The zero locus of the moment map \(\mu^{-1}(0)\) is then given by
\[ xy - yx + ij = 0 \in \mathrm{End}(k^n). \]
We can deduce that on \(\mu^{-1}(0)\), we have
\[ ji = \mathrm{tr}(ji)= \mathrm{tr}(ij) = -\mathrm{tr}(xy-yx) = 0. \]
We know that if \(j=0\), then the equivalence classes of \((x,y,i)\) in which \(i\) give a cyclic vector is in bijection with \(\mathrm{Hilb}^n(\mathbb{C}^2)\). It turns out that this can be deduced considering only a subset of such representation space, i.e. by choosing a suitable stability parameter \(\theta\) and consider the semistable locus.
Setting Stability Parameter Eliminates \(j\)
Let \(\theta = (n, -1)\), then our representation \((x,y,i,j)\) is semi-stable iff it does not contain any subrepresentation of dimension \((1,m)\) for \(m<n\), this means if image of \(i\) is non-zero, it has to generate the entire \(k^n\) together with \(x\) and \(y\).
With this generation requirement, \(j\) will be forced to be \(0\). We can see this by considering any closed path \(j P i\) where \(P\) is any (non-zero) number of products of \(x,y\), we have
\[\begin{align*} j P i &= \mathrm{tr}(j P i) \\ &= \mathrm{tr}(i j P) \\ &= -\mathrm{tr}([x,y]P) \\ &= 0, \end{align*}\]
since we can always put \(x\), \(y\) to be simultaneously upper triangular form, by a result in linear algebra and using the fact that \([x,y]=-ij\) has rank \(1\).
This means \(j=0\) since \(i\) is cyclic vector and thus it is zero on the entire \(k^n\).