Correct. $\displaystyle{16}{x}{{z}}^{{2}}\usepackage{cmbright}\Huge\$ and $\displaystyle-{5}{x}{{z}}^{{2}}\usepackage{cmbright}\Huge\$ have the same variables and the same exponents, so they are like terms. $\displaystyle-{3}{z}\usepackage{cmbright}\Huge\$ and $\displaystyle-{3}{{z}}^{{2}}\usepackage{cmbright}\Huge\$ have the same variable but different exponents, so they are unlike terms.