Here is a way to show that $Y$ is irreducible with little computation: note that since $Y$ is closed (as the set of zeros of two polynomials) we have $Y=Z(I(Y))$, where $I(Y)$ is the ideal of $Y$. The only thing to check is that $I(Y)$, is prime. But it is easy to see that $I(Y)$ is the kernel of a surjective ring homomorphism $\phi:k[x,y,z]\to k[t]$, namely the homomorphism such that $\phi(x)=t,\phi(y)=t^2,\phi(z)=t^3$. Let's check this in detail, even though this is a bit tautological: To say that $p\in k[x,y,z]$ belongs to $\ker\phi$ means that the polynomial $p(t,t^2,t^3)\in k[t]$ is the zero polynomial. This implies, obviously, that for all $t\in k$ we have $p(t,t^2,t^3)=0$, i.e. that $p\in I(Y)$. Conversely, if $p\in I(Y)$ then the polynomial $\phi(p)=p(t,t^2,t^3)\in k[t]$ has any $t\in k$ as a root, which means that $\phi(p)=0$. Hence $I(Y)=\ker\phi$.
It follows that the ring $A(Y)=k[x,y,z]/I(Y)$ is isomorphic to $k[t]$, which is integral, hence $I(Y)$ is prime, and $Y$ is an affine variety isomorphic to the affine line. Q.E.D.
Of course the ideal $I(Y)$ is the radical of $(x^2-y,x^3-z)$, and the above arguments show that in fact $I(Y)=(x^2-y,x^3-z)$, but the computation of $I(Y)$ is not needed.