I am looking for all irreducible polynomials of degree 5 in $\mathbb{F}_{17}$ with have the form h(y) = $y^5+C$.
I think there aren't any irreducible polynomials of this form because for every C I can find an element of $\mathbb{F}_{17}$ as a root.
What would be the fastest way to find irreducible polynomials if the form would be h(y) = $y^5$+...
You are correct that there no irreducible polynomials of the form $h(y)=y^5+C$ over $\Bbb F_{17}$ because every such polynomial has a root in $\Bbb F_{17}$. An easy way to see this is to notice that $\Bbb F_{17}^*$ is a cyclic group of order 16, which is relatively prime to 5, so that $y \mapsto y^5$ is an automorphism of $\Bbb F_{17}^*$. This means that $y \mapsto y^5$ is a bijection from $\Bbb F_{17}$ to itself, so for any $C$, there exists some $y\in\Bbb F_{17}$ such that $y^5=-C$.
Now, to find an irreducible polynomial of degree 5, a reasonably fast way to do it is to simply pick monic polynomials of degree 5 at random and test them until you find one that is irreducible; about $\frac15$th of them are irreducible (see below), so you probably won't have to try too many.
The question then becomes: how do we test a degree 5 polynomial $f$ for irreducibility? Well, if $f$ is reducible then it must have either a quadratic or a linear factor. Now, the polynomial $x^{17^2}-x=x^{289}-x$ is the product of all irreducible linear and quadratic polynomials over $\Bbb F_{17}$, so we can test for a linear or quadratic factor by checking if $\text{gcd}(f,x^{289}-x) \neq 1$. We can make this easier by recognizing that if we replace $x^{289}-x$ by its reduction mod $f$ then it does not change the gcd with $f$, and we can do this by computing the power $x^{289}$ mod $f$ using repeated squaring. If you're trying to do this by hand, this is still going to require quite a bit of calculation, but on a computer it is very fast and also scales nicely to larger finite fields and higher-degree polynomials (Look up the Ben-Or irreducibility test).
(Proof of the fact that about $\frac15$th of monic degree 5 polynomials are irreducible: $x^{289^5}-x$ factors over $\Bbb F_{17}$ as the product of all of the irreducible polynomials of degree 1 or 5; there are 17 irreducible polynomials of degree 1, so by considering the degree of $x^{17^5}-x$ we conclude that there are $\frac{17^5-17}{5}$ irreducible polynomials of degree 5. So the fraction of all monic degree 5 polynomials that are irreducible is $\frac1{17^5}\cdot \frac{17^5-17}{5} = \frac{1-17^{-4}}5$.)
The unit group has 16 elements, hence any element $a$ has order relatively prime to $5$. In particular $a$ and $a^5$ have the same order in the cyclic group generated by $a$. We obtain $a \in \langle a \rangle = \langle a^5 \rangle$.
Edit There would be the following approch to easily find such an irreducible polynomial: If $17$ would have order $5$ mod $11$, we would deduce that there exists a $11$th root of unity in $\mathbb F_{17^5}$. This would mean that $\frac{x^{11}-1}{x-1}$ factors into two irreducibles of degree $5$. Unfortunately $17$ is a primitive root modulo $11$, hence $\frac{x^{11}-1}{x-1}$ is irreducible itself.
So this does not work here (though there were some chances a priori...), but of course there are many situations, where this approach works well.
