Counting the $\operatorname{XOR}$ of an Element over the Finite Field $\operatorname{GF}(2^q)$
Let $f=x^q+t$ be an irreducible polynomial, then


  $$\alpha\cdot \beta= (\alpha_{q-1},\dots,\alpha_0)\cdot(\beta_{q-1},\dots,\beta_0):
\bigoplus_{i=0,\alpha_i\neq0}^{q-1}\big( \alpha_i\cdot(\beta&lt;&lt;i) \oplus (\bigoplus_{j=1}^{i} \beta_{q-j}\cdot x^{j-1}\cdot t)\big).$$


With this formula, you can easily calculate XOR count of $\alpha$. Although there may be a simpler solution to answer your question.
Note that in this formula, several $\beta_ k$ may be xor with each other, which should not be taken into account in the XOR count calculation. For this, a $q\times q$ matrix can be used for implementation.

Edit: As an example, let $f=x^8+x^4+x^3+x+1=x^8+t$. Then
$$(x+1)\cdot \beta=\alpha_0\cdot (\beta_7,\dots,\beta_0)\oplus \alpha_1\cdot (\beta_6,\dots,\beta_0,0)\oplus \beta_7 \cdot t=(\beta_7,\dots,\beta_0)\oplus (\beta_6,\dots,\beta_0,0)\oplus (0,0,0,\beta_7,\beta_7,0,\beta_7,\beta_7)=(\beta_7\oplus \beta_6,\beta_6\oplus \beta_5,\beta_5\oplus \beta_4,\beta_4\oplus \beta_3\oplus \beta_7,\beta_3\oplus \beta_2\oplus \beta_7,\beta_2\oplus \beta_1,\beta_1\oplus \beta_0\oplus \beta_7,\beta_0\oplus \beta_7).$$
So, the XOR count of $x+1$ is $11$.