Definition. $x$ is called involution if it has order $2$.
Problem. I have to count maximal number of pairwise commuting involutions in $Sp(2n, F)$, $\text{char}F \neq 2$.
Attempt. Firstly, let's remember that pairwise commuting matrices are simultaneous diagonalizable so in some basis they will be diagonal with $-1$ and $1$ on the diagonal. Unfortunately, $\Omega$ (which is used in definition of $Sp(2n, F)$) is also changed. Changing basis we can make $$ \Omega = \begin{pmatrix} \Omega_1 &0\\ 0 &\Omega_2 \end{pmatrix}, $$ where $\Omega_1$ and $\Omega_2$ are even dimensionality antisymmetric matrices and our involutions are still in diagonal form. Now I'm stuck and don't know how to proceed.
The largest elementary abelian $2$-subgroup of ${\rm Sp}(2n,F)$ has order $2^n$, so the answer is $2^n-1$.
To see this, let $E$ be an elementary abelian $2$-subgroup. Then $E$ fixes some $1$-dimensional subspace $\langle v \rangle$ of $F^{2n}$, and hence it stabilizes the subspace $\langle v \rangle^\perp$, which has dimension $2n-1$.
Now, since $E$ is completely reducible, it also fixes a $1$-dimensional subspace $\langle w \rangle$ with $w \in F^{2n} \setminus \langle v \rangle^\perp$, and in order to preserve the form on $\langle v,w \rangle$, the induced action of $E$ on this $2$-dimensional subspace has order at most $2$. Now $E$ preserves $\langle v,w \rangle^\perp$ of dimension $2n-2$, and by induction its induced action on that subspace has order at most $2^{n-1}$, so $|E| \le 2^n$.
There exist elementary abelian subgroups of order $2^n$ as subgroups of ${\rm Sp}(2,F)^n \le {\rm Sp}(2n,F)$.