Is a polynomial $y^n+y^{n-1}-x^m-x^{m-1}$ irreducible in $\Bbb Z[x,y]$?

The question is in the title, $n>m\ge 2$ are integers. All text below is the context.

Two weeks ago user759001 asked on integer solutions $x>y\ge 2$ of a Diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ for integers $m,n\geq 2$. The only known solutions are $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$. User2020201 showed that $m<n$. I conjectured that there are no solutions when $m|n$ and proved the conjecture in particual cases (when $(m,n)$ is $(2,6)$, $(3,19)$, or $(4,12)$. Also I guess I have a proof when $n=2m$), see this answer.

According to [G], Diophantine equations with two variables of degree greater than two have infinitely many (integer) solutions only in very rare cases. In particular, by a special and very complicated method K. Zigel’ (Siegel?) showed the following

Theorem. Let $P(x,y)$ be an irreducible polynomial of two variables with integer coefficients of a total degree greater than two (that is, $P(x,y)$ contains a monomial $ax^ky^s$, where $k+s>2$). (The irreducibility of $P(x,y)$ means that it cannot be represented as a product of two non-constant polynomials with integer coefficients). If an equation $P(x,y)=0$ has infinitely many integer solutions $(x,y)$ then there exist an integer $r$ and integers $a_i$, $b_i$ for each $-r\le i\le r$ such that if in the equation $P(x,y)=0$ we make a substitution $x=\sum_{i=-r}^r a_it^i$ and $y=\sum_{i=-r}^r b_it^i$ then we obtain an identity.

In order to apply this theorem to user759001’s equation for fixed $n>m>2$ we need irreducibility of the polynomial $y^n+y^{n-1}-x^m-x^{m-1}$. It looks plausible and easy to show, but, unfortunately, I am not a specialist in factorization of multivariable polynomials, so I decided to ask MSE community for help. Thanks.

References

[G] Gel’fand A.O. Solutions of equations in integer numbers, 3rd edn., Moscow, Nauka, 1978, in Russian.

Edit. I have posted now a complete answer...

Since $P=y^n+y^{n-1}-x^m-x^{m-1}$ is monic in $y$ and $\mathbb{Z}[x]$ is a UFD with quotient field $\mathbb{Q}(x)$, it is enough to prove irreducibility over $\mathbb{Q}(x)$.

The Newton polygon wrt to the $x$-adic valuation has vertices $(0,m), (n-1,0), (n,0).$ There is no other intersection points with $\mathbb{Z}^2$, menaning that a non trivial factor of $P$ has degree $n-1-0=n-1$ or $n-(n-1)=1$ over $\mathbb{Q}((x))$.

In particular, a non trivial factor of $P$ (if any) over $\mathbb{Q}(x)$ has degree $n-1$ or $1$. In both cases, $P$ has a root in $\mathbb{Q}(x)$.

Let $r/s\in\mathbb{Q}(x)$ a root of $P$, where $r,s\in \mathbb{Q}[x]$ are coprime. Without loss of generality, one may assume that $s$ is monic. Since $r^n/s^n+ r^{n-1}/s^{n-1}-x^m-x^{m-1}=0$, we get $r^n=s\times$ some polynomial. Hence $s\mid r$, and thus $s=1$ (since $s$ is monic and coprime to $r$).

So $r$ is a root of $P$, and $r^n+r^{n-1}=r^{n-1}(r+1)=x^m+x^{m-1}=x^{m-1}(x+1)$. In particular, $r$ is not constant.

Note that $r$ and $r+1$ are coprime, so they have different irreducible factors. The only possible irreducible factors are $x$ and $x+1$. If $r=cx^k, c\in \mathbb{Q}^\times,k\geq 1$, then $(n-1)k=m-1$. This is not possible since $n>m$ and $k\geq 1$. Thus $r=c (x+1)^k$. Comparing $x+1$-adic valuations yields this time $(n-1)k=1$, so $k=1$ and $n=2$, which is also excluded since $n\geq 3$.