= Solution
The operators associated with the two angular momenta commute with one another, and each squared <angular momentum operator> commutes with every component of that same <angular momentum>. Two useful identities are
$$
J_z=J_{1z}+J_{2z},\qquad
\mathbf J_1\cdot\mathbf J_2=\frac12(J^2-J_1^2-J_2^2).
$$
Consequently the two maximal commuting subsets of the displayed collection are
$$
\boxed{\{J_1^2,J_2^2,J_{1z},J_{2z},J_z\}},\qquad
\boxed{\{J_1^2,J_2^2,J^2,J_z,\mathbf J_1\cdot\mathbf J_2\}}.
$$
The first set gives the uncoupled <eigenbasis> $|j_1m_1\rangle\otimes|j_2m_2\rangle$. Its $J_z$ <eigenvalue> is already fixed by $m=m_1+m_2$. The second gives the coupled <eigenbasis> $|j_1j_2;jm\rangle$ for the <addition of angular momentum>. Its scalar-product <eigenvalue> is already fixed by $j,j_1,j_2$. Thus an independent <complete set of commuting observables> uses four operators in either case; literal maximality within the question's list includes the dependent fifth operator.
For generic nonzero angular momenta, no further listed operator can be added. In particular,
$$
[J_{1z},\mathbf J_1\cdot\mathbf J_2]
=i\hbar(J_{1y}J_{2x}-J_{1x}J_{2y})
$$
is generally nonzero, and $[J_{1z},J^2]=2[J_{1z},\mathbf J_1\cdot\mathbf J_2]$. The same obstruction applies to $J_{2z}$. Their sum $J_z$ commutes because the two <commutators> cancel.
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