$AB-BA=I$ having no solutions

Let a $n\times n$ matrix. Take trace of both sides $$\operatorname{trace}(AB-BA)= \operatorname{trace}(I)\Rightarrow \operatorname{trace}(AB)- \operatorname{trace}(BA) =n\Rightarrow0=n$$

As you should have known by now, for real matrices, the equation $AB-BA=I$ has no solution because the LHS has zero trace but the RHS is not traceless. The same conclusion holds for complex matrices. For other fields, an in-depth discussion can be found in the answers to a related question. In particular, it has been proven that a square matrix $M$ is a commutator (i.e. $M=AB-BA$ for some square matrices $A$ and $B$) if and only if $M$ is traceless:

A.A. Albert and Benjamin Muckenhoupt (1957), "On matrices of trace zeros". Michigan Math. J., 4(1):1-3.

It follows that $AB-BA=I_n$ has a solution if and only if $I_n$ is traceless over the underlying field.

Another way is to use the fact that $AB$ and $BA$ have the same set of eigenvalues.

Rewrite the equation as $AB=BA+I$, then it follows that $\lambda$ is an eigenvalue of $AB$ iff $\lambda$ is an eigenvalue of $BA+I$, or equivalently, $\lambda-1$ is an eigenvalue of $BA$.

Repeating the argument, we can see that if $\lambda$ is an eigenvalue of $AB$, then for any postive integer $n$, $\lambda-n$ is an eigenvalue of $AB$, this means $AB$ have infinitely many distinct eigenvalues, a contradiction.