First of all, you're mixing $\mathcal{Z}$-transform and discrete-time Fourier transform (DTFT) here. Just because
$$\mathcal{Z}\big\{u[n]\big\}=\frac{1}{1-z^{-1}}\tag{1}$$
doesn't mean that
$$\textrm{DTFT}\big\{u[n]\big\}\stackrel{?}{=}\frac{1}{1-e^{-j\omega}}\tag{2}$$
The correct expression is
$$\textrm{DTFT}\big\{u[n]\big\}=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{3}$$
[Cf. this question and its answers.]
Apart from that, you are right and your professor is wrong. Your professor's expression should look like this:
$$-e^{-3j\omega}\frac{1}{1-e^{-j\omega}}\tag{4}$$
(note the negative sign in the exponent), because he didn't replace $n$ by $n+3$ but by $n-3$, hence the multiplication with $e^{-3j\omega}$.
Now it's straightforward to show that his (corrected) solution and yours are identical:
$$-e^{-3j\omega}\frac{1}{1-e^{-j\omega}}=-e^{-2j\omega}\frac{1}{e^{j\omega}-1}=e^{-2j\omega}\frac{1}{1-e^{j\omega}}\tag{5}$$$$-\frac{e^{-3j\omega}}{1-e^{-j\omega}}=-\frac{e^{-2j\omega}}{e^{j\omega}-1}=\frac{e^{-2j\omega}}{1-e^{j\omega}}\tag{5}$$
With $(3)$ we have
$$\textrm{DTFT}\big\{u[-n+2]\big\}=\pi\delta(\omega)+\frac{e^{-2j\omega}}{1-e^{j\omega}}\tag{6}$$