= Solution
An <eigenvalue> of $C$ is a scalar $\lambda$ for which a nonzero vector $v$ satisfies $Cv=\lambda v$. Equivalently, $\det(\lambda I-C)=0$, because the homogeneous linear system has a nontrivial solution exactly when its coefficient <matrix> is singular. This <characteristic polynomial> is the monic quadratic
$$
\lambda^2-(\operatorname{tr}C)\lambda+\det C.
$$
A nonzero quadratic has at most two distinct roots, proving the <eigenvalue> bound. <Eigenvalues> and diagonalization are taken over $\mathbb C$, also for a <matrix> with real entries.
For the <matrix commutator>, the cyclic <trace> identity follows directly from its entries:
$$
\operatorname{tr}(AB)=\sum_{i,j}A_{ij}B_{ji}
=\sum_{i,j}B_{ij}A_{ji}=\operatorname{tr}(BA).
$$
Hence \b[$\operatorname{tr}[A,B]=0$]. Factoring the <characteristic polynomial> as $(\lambda-\lambda_1)(\lambda-\lambda_2)$ and comparing the coefficient of $\lambda$ gives \b[$\operatorname{tr}C=\lambda_1+\lambda_2$], with the <eigenvalues> counted with algebraic multiplicity. If both <eigenvalues> are zero, <trace> and <determinant> are zero. The <Cayley-Hamilton theorem> then gives \b[$C^2=0$]; no diagonalizability assumption is needed.
Now put $M=[A,B]$. Since its <trace> is zero, its <characteristic polynomial> is $\lambda^2+\det M$ and the <scalar square of a traceless two-by-two matrix> identity gives
$$
\boxed{[A,B]^2=\alpha I,\qquad \alpha=-\det[A,B]
=\frac12\operatorname{tr}([A,B]^2).}
$$
This proves the scalar-square assertion for both real and complex entries. If $\det M=0$, it gives $M^2=0$. If $\det M\ne0$, the two <eigenvalues> are the distinct nonzero numbers $\pm\sqrt{-\det M}$. Their eigenvectors are <linearly independent>: a nonzero vector cannot satisfy two different <eigenvalue> equations. They form a <basis>, making $M$ a <diagonalizable matrix>. Thus \b[either the commutator is diagonalizable over $\mathbb C$ or its square is zero]. These alternatives need not be disjoint, since the zero <matrix> has both properties. Over $\mathbb R$ the diagonalization assertion would require real <eigenvalues>; complex diagonalization is essential to the unrestricted statement.
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