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Post question's edit:
You want to prove that the polynomial is reducible in $\Bbb Z[x]$. By (the second) Gauss's lemma, if it is reducible in $\Bbb Q[x]$, then it also is in $\Bbb Z[x]$. Since its degree is $3$, it is reducible (over $\Bbb Q$) if, and only if, there is a rational root. Now note that in case Eistein's criterion doesn't work, you still won't know wether it is reducible or not, so that shouldn't be the way to go. You want to find a rational root. Try finding roots using the rational root theorem.
Something else you can try is to factor your polynomial. If you can't do this, you can always put the polynomial as input in a very well known on-line software.
Best Answer
We apply the Eisenstein Criterion as follows. We view $\mathbb{C}[x, y]$ as a polynomial ring $R[x]$ in one variable, where we set $R=\mathbb{C}[y]$.
So the polynomial $f(x)\in R[x]$ is given by $f(x) = x^{m} + y^{n}-1$. In order to apply the Eisenstein Criterion to the element $y-1\in R$, we need check that $y-1$ does not divide the leading coefficient of $f(x)$, which is $1$ in this case; and $y-1$ divides $y^{n}-1$ to the first power, that is, $y-1 \mid y^{n}-1$ but $(y-1)^2 \not\mid y^{n}-1$.
Since $y^{n}-1=(y-1)(y^{n-1}+y^{n-2}+\cdots + y+1)$, it is clear that $y-1\mid y^{n}-1$. In order to show $(y-1)^{2}\not\mid y^{n}-1$, we need that $y-1\not\mid y^{n-1}+y^{n-2}+\cdots + y + 1$. One way to see this is to note that $1$ is a root of $y-1$, but not a root of $y^{n-1}+y^{n-2}+\cdots+y+1$.