Irreducibility of $x^4-5$ over $\mathbb{Z}_{17}$ Obviously, it'd be hard to try all the $17$ elements to see if there is some root, and even if there is none, it'd be necessary to verify if it can't be factored into two irreducible 2 degree polynomials. The Eisenstein criterion will just be able to say about irreducibility on $\mathbb{Q}[x]$.
What can I do?
I must also verify it for the polynomials $x^3-5$ and $x^4+7$, both over $\mathbb{Z}_{17}$.

Assuming you mean the finite field $\mathbb F_{17}$ as opposed to the $17$-adic integers $\mathbb Z_{17}$, here is a method you can use to check for quadratic factors:
There is a unique field extension $\mathbb F_{17^2}$ of $\mathbb F_{17}$ or degree $2$. Thus, every irreducible quadratic polynomial defines the same field extension. Therefore, $x^4 - 5$ has no quadratic factors if and only if it does not have a root in $\mathbb F_{17^2}$.
The order of $5$ in $\mathbb F_{17}^\times$ is at most $16$ by Fermat's little theorem. On the other hand, if the order were $8$ or lower, then $5$ would be a square (because $\mathbb F_{17}^\times$ is cyclic). This is impossibly by user236182's answer. Therefore, $5$ is a primitive unit in $\mathbb F_{17}$.
Since the group $\mathbb F_{17^2}^\times$ is cyclic of order $17^2 - 1 = 288 = 32 \cdot 9$, and $5$ has order $16$, we conclude that $5$ cannot be a fourth power in $\mathbb F_{17^2}^\times$ (for otherwise the order would divide $\frac{288}{4} = 8 \cdot 9$, which it clearly doesn't).
Thus, the polynomial $x^4 - 5$ is irreducible over $\mathbb F_{17}$.
