First of all, am I right in stating $I=(2x^3-x^2+2x+5)=(2x^3+4x^2+2x)=(x^3+2x^2+x)=\{\alpha(x^3+2x^2+x)\vert \alpha\in\mathbb{Z}_5\}?$ In this case $I$ is proper.
Also, in the solution my professor wrote $I=(x)$ but I can't see how he got that. Analogously, in $\mathbb{Z}_3[x]$ how is $I=(x^2+1)$? Using the (maybe flawed) reasoning above and Fermat I can only get $I=(2x^2+x+2)$.
First, 5 is the same as 0 in $\Bbb Z_5$. With this in hand, it becomes obvious that both generators are multiples of $x$, and therefore $I$ is contained in the principal ideal $(x)$ (which is already a proper ideal, so $I$ is a proper ideal as well).
It is also trivial to see that in fact $I$ is equal to $(x)$, but why does that matter if the problem was to prove that $I$ is a proper ideal? The difference of the two generators is $x^2$, so $I$ contains $x^2$ and therefore $x^3$ as well, so it also contains $x$ because it contains $x^3+x$. So you get inclusions in both directions.
In $\Bbb Z_3$ 5 = -1, so the difference of the two generators is $x^2+1$. Both generators are divisible by $x^2+1$, so again you get $I=(x^2+1)$ by inclusion in both directions.