Let $k$ be a field of characteristic zero, and denote the first Weyl algebra by $A_1(k)$, namely, $A_1(k)$ is the associative, non-commutative $k$-akgebra generated by $x$ and $y$ subject to the relation $[y,x]=yx-xy=1$.
Given an element $A \in A_1(k)$, call an element $B \in A_1(k)$ a Dixmier mate of $A$, if $[A,B] \in A_1(k)^{\times}=k^{\times}$; in this case, call $(A,B)$ a Dixmier pair.
Recall that:
(1) Any $a,b,c \in A_1(k)$ satisfy: $[ab,c]=a[b,c]+[a,c]b$; indeed, $[ab,c]=(ab)c-c(ab)=abc-cab=abc-acb+acb-cab=(abc-acb)+(acb-cab)=a(bc-cb)+(ac-ca)b=a[b,c]+[a,c]b$.
(2) For $t(y) \in k[y]$, we have: $[t(y),x^i]=i x^{i-1}t'(y)+\binom{i}{2} x^{i-2}t''(y)+\dots$.
I wish to show that $(x+y)^2$ does not have a Dixmier mate.
Denote $B=\sum b_{ij} x^iy^j$. Then if $[(x+y)^2,B] \in k^{\times}$, then by (1) we get $(x+y)[x+y,B]+[x+y,B](x+y)=[(x+y)^2,B] \in k^{\times}$.
With the help of (1) and (2), I have further computed $[x,B]$ and $[y,B]$ (perhaps I will add my computations later), but was too lazy to further compute $(x+y)[x,B]$, $(x+y)[y,B]$, $[x,B](x+y)$ and $[y,B](x+y)$.
Is there a shorter, more elegant way than the one I suggest, to show that $(x+y)^2$ does not have a Dixmier mate?
Remark: The commutative analog is easy: Otherwise, there exists $B \in k[x,y]$ such that $Jac((x+y)^2,B) \in k^{\times}$. But, $Jac((x+y)^2,B)=(2x+2y)B_y-(2x+2y)B_x=(2x+2y)(B_y-B_x)$; hence, $Jac((x+y)^2,B)|(0,0)=0$, which contradicts $Jac((x+y)^2,B) \in k^{\times}$.
If I am not wrong, substituing $(x,y)$ for $(0,0)$ is only possible in the commutative case (by applying an appropriate uviversal property).
In general, if $1=[A^2,B]$, then by Lemma 2.4 of "Sur de algebres de Weyl" of Dixmier you have $$ 1=\mathcal{l}_{11}(1)=\mathcal{l}_{11}(A[A,B]+[A,B]A)=2 \mathcal{l}_{11}(A[A,B]), $$ in particular $v_{11}(A)=\deg(A)=0$ which implies $A\in K^{*}$ and leads to the contradiction $[A^2,B]=0$.