Consider quintic polynomials in the form $$f=T^{5}+2^{q}3^{4k+1}5^{4r+1}T+2^{5q+2}3^{5k+1}5^{5r}\quad q,k,r\in \Bbb N \tag{1}$$ whose discriminant is $\Delta=\left(2^{5+10q}3^{2+10k}5^{2+10r}\right)^{2}5\notin \Bbb Z^2$, not a square.
For example,$(q,k,r)=(0,0,0)$ gives $f=T^5=15T+12$. This family has unique real root, $f'>0$, and the splitting field contains $\Bbb Q(\sqrt 5)=\Bbb Q(\zeta_5)$). It is build by considering reduced quintics $T^5+AT+B$ whose sextic resolvent $R=\prod_{g\in S_5/F_{20}}(T-g\phi)$ has zero as rational root ($256A^{6}-9375AB^{4}=0$ gives this family), where the resolvent $\phi(x_0,\dots,x_4)=\sum x_{\sigma(i)}x_{\sigma(j)}x_{\sigma((i+j)/2}$ is stablized by the affine linear group $F_{20}$ of order $20$, as explained in Dummit's paper on solvable quintics, cf Wiki Solvable Quintics)
EDIT: the question was to show that the polynomials (1) are irreducible. Servaes proposed the transformation $3^{-5k}f(3^kT)=T^5 +2^{4q}\cdot 3\cdot 5^{4r+1}T+2^{5q+2}\cdot 3\cdot 5^{5r}$ that shows the polynomial is $3$-Eisenstein, so irreducible.