The linear operator $B \to AB + BA$ corresponds to the $4 \times 4$ matrix $ M = \pmatrix{2a & b & c & 0\cr c & a+d & 0 & c\cr b & 0 & a+d & b\cr 0 & b & c & 2d\cr}$ using the basis $\left(\pmatrix{1 & 0\cr 0 & 0\cr}, \pmatrix{0 & 1\cr 0 & 0\cr},\pmatrix{0 & 0\cr 1 & 0\cr}, \pmatrix{0 & 0\cr 0 & 1\cr}\right)$ of the $2 \times 2$ matrices. This has determinant $-4\,bc{d}^{2}-8\,bcad-4\,bc{a}^{2}+4\,a{d}^{3}+8\,{d}^{2}{a}^{2}+4\,d{ a}^{3}=4\, \left( a+d \right) ^{2} \left( ad-bc \right) $ so there are nonzero solutions for $B$ if and only if either $a+d=0$ or $ad-bc=0$. If $a+d=0$, substituting $d = -a$ in $M$ we get $ \pmatrix{2a & b & c & 0\cr c & 0 & 0 & c\cr b & 0 & 0 & b\cr 0 & b & c & -2a\cr}$ As long as $a,b,c$ are not all $0$, this has rank $2$, so there is a $2$-dimensional linear space of $B$ for which $AB + BA = 0$ in this case. On the other hand, if $ad - bc = 0$ but $a+d \ne 0$ the matrix should have rank $3$ and we should only have a $1$-dimensional linear space of $B$: in fact the $(1,1)$, $(2,3)$, $(3,2)$ and $(4,4)$ entries of the classical adjoint matrix are $2(d^2 + ad-bc)(a+d)$, ($2 c^2 (a+d)$, $2 b^2 (a+d)$ and $2 (a^2 + ad-bc)(a+d)$ respectively: if $ad-bc=0$, $a+d \ne 0 $ and the rank is less than $3$, all these would be $0$ which implies $a=b=c=d=0$.
EDIT: In the $3 \times 3$ case, the determinant of the $9 \times 9$ matrix is $\eqalign{8\, &\left( a_{{3,3}}a_{{1,1}}a_{{2,2}}-a_{{2,3}}a_{{3,2}}a_{{1,1}}+a_{ {1,3}}a_{{2,1}}a_{{3,2}}-a_{{3,3}}a_{{1,2}}a_{{2,1}}+a_{{2,3}}a_{{1,2} }a_{{3,1}}-a_{{1,3}}a_{{3,1}}a_{{2,2}} \right)\cr &( -a_{{2,3}}a_{{1 ,2}}a_{{3,1}}-a_{{1,3}}a_{{3,1}}a_{{3,3}}-a_{{1,2}}a_{{2,1}}a_{{2,2}}+ {a_{{3,3}}}^{2}a_{{1,1}}+a_{{1,1}}{a_{{2,2}}}^{2}-a_{{1,3}}a_{{1,1}}a_ {{3,1}}-a_{{1,1}}a_{{1,2}}a_{{2,1}}+{a_{{3,3}}}^{2}a_{{2,2}}\cr&+2\,a_{{3, 3}}a_{{1,1}}a_{{2,2}}+a_{{3,3}}{a_{{2,2}}}^{2}-a_{{2,3}}a_{{3,2}}a_{{3 ,3}}-a_{{1,3}}a_{{2,1}}a_{{3,2}}-a_{{2,3}}a_{{3,2}}a_{{2,2}}+{a_{{1,1} }}^{2}a_{{2,2}}+a_{{3,3}}{a_{{1,1}}}^{2} ) ^{2}\cr} $ It looks to me like there is a $1$-dimensional solution space when one of these factors is $0$ and a $2$-dimensional space when the other is $0$. Then there are quite a number of cases where both factors are $0$, some of which lead to $3$-dimensional spaces.