Let's start with Fermat equation with the lowest power, $x^3 + y^3 = z^3$. Now let's set $y = x + a, z = x + b$ with $b > a$ and $a,b$ integers. then the equation becomes
$$x^3 + (3a-3b)x^2 + (3a^2-3b^2)x + (a^3-b^3) = 0\tag 1$$
Since we know that Fermat's equation does not have a solution, we know that the above cubic will never admit integer solutions. The question is how to prove it.
Edit: complete restart. This is posted with the assumption that it is likely to have a fallacy, since it is too simple and obvious to have been missed for $350$ years, but with the hope that the methods presented may be useful.
Let $n=2m+1$ be an odd prime, assume that $x,y,z,a,b\in\Bbb Z$ and consider the solutions in $x$ of the following polynomial:
$$x^n+y^n=z^n=x^n+(x+a)^n=(x+b)^n\implies x^n+\sum_{i=0}^{n-1}{n\choose i}\left(a^{n-i}-b^{n-i}\right)x^i=0\tag 1$$
The solutions in $x$ become the symmetric polynomials of degree $1$ to $n$ of $n$ distinct solutions. Let these solutions be labeled $u_k, k\in[0,n-1]$ and let these polynomials take their values as follows:
$$\sum_{k=0}^{n-1}u_k = n(a-b)\tag {$p_1$}$$ $$\sum_{k=0,}^{n-2}\sum_{\ell=k+1}^{n-1}u_ku_\ell = {n\choose 2}(a^2-b^2)={n\choose 2}(a-b)(a+b)\tag {$p_2$}$$ $$\sum_{k=0,}^{n-3}\sum_{\ell=k+1,}^{n-2}\sum_{m=\ell+1}^{n-1}u_ku_\ell u_m = {n\choose 3}(a^3-b^3)={n\choose 3}(a-b)(a^2+ab+b^2)\tag {$p_3$}$$ $$\vdots$$ $$\prod_{k=0}^{n-1}u_k = a^n-b^n\tag{$p_n$}=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^i$$
Begin by claiming that at least one integer solution exists for $x$ in $(1)$; let this solution be the one labeled $u_0$ above. Each of the symmetric polynomials $(p_1),\dots,(p_n)$ above are integers on the RHS and must therefore be integers on the LHS as well. Since $u_0$ is claimed to be an integer, we must also have $\sum_{k=1}^{n-1}u_k$ be an integer in $(p_1)$. By the same token, $u_0\sum_{k=1}^{n-1}u_k$ is an integer in $(p_2)$ and therefore $\sum_{k=1,}^{n-2}\sum_{\ell=k+1}^{n-1}u_ku_\ell$ is also an integer in $(p_2)$. This process continues through the rest of these polynomials until we see that $\prod_{k=1}^{n-1}u_k$ must be an integer in both $(p_{n-1})$ and $(p_n)$.
Next, consider the divisibility facet of these polynomials. Each has a factor $a-b$ as demonstrated. Note this means that $a-b\mid x$ for any integer solution $x$. Each of the polynomials $(p_1),\dots,(p_{n-1})$ also has a factor $n$ since $n$ is prime and $\forall i\in[1,n-1],n\mid{n\choose i}$. Let $f$ be a prime factor of $a-b$. From $(p_n)$ we see that either $f\mid u_0$ or $f\mid\prod_{k=1}^{n-1}u_k$. In the symmetric polynomials $(p_1),\dots,(p_{n-1})$ there is more than one term and at least one term in each that does not involve $u_0$. Denote the portion of a given polynomial $(p_j)$ which contains only terms which have $u_0$ in them as $(p_j)[u_0]$. Then the portion of that polynomial which contains only terms which do not have $u_0$ in them is $(p_j)-(p_j)[u_0]$.
Now the process used to determine that $\forall j\in[1,n-1], (p_j)-(p_j)[u_0]\in\Bbb Z$ is repeated, except this time we are considering the given prime factor $f$. In particular, assume that $f\mid\prod_{k=1}^{n-1}u_k$. Then we have that $f\mid (p_{n-1})$ and also that $f\mid (p_{n-1})-(p_{n-1})[u_0]$ and therefore $f\mid (p_{n-1})[u_0] = u_0\sum_{k=1,}^{n-1}\prod_{\ell\neq k}u_\ell$. But $\sum_{k=1,}^{n-1}\prod_{\ell\neq k}u_\ell=(p_{n-2})-(p_{n-2})[u_0]$ and therefore we apply the same divisibility constraint to see that $f\mid (p_{n-2})[u_0]$. This process continues uninterrupted until we arrive at $f\mid u_0$ in $(p_1)$. We knew previously that either $f\mid u_0$ or $f\mid\prod_{k=1}^{n-1}u_k$; now we know that $f\mid u_0$, so we can start at $(p_1)$ and note that this means that $f\mid (p_1)-(p_1)[u_0]$, which further means that $f\mid (p_2)[u_0]$ and thus that $f\mid (p_2)-(p_2)[u_0]$, and again the process continues until we get back to $(p_n)$ and find that $f^2\mid (p_n)=(a^n-b^n)$. At this point, we either have $f^2\mid (a-b)$ or $f\mid\tfrac{a^n-b^n}{a-b}$. In the first case, we can jump to $(p_2)$ and note that $(p_2)=u_0((p_1)-(p_1)[u_0])+(p_2)-(p_2)[u_0]=n(a-b)(a+b)$ which means $f^2\mid (p_2), f^2\mid (p_2)[u_0]$ and therefore $f^2\mid (p_2)-(p_2)[u_0]$. But then this tracks back up to $(p_n)$ as $f^n\mid (p_n)$, and any further such factors demonstrate the same tendency, which leads to $C^n\mid a-b$ for some $C\in\Bbb Z$ unless we allow $f\mid \tfrac{a^n-b^n}{a-b}$. First, consider the branch assumption that $f\mid\tfrac{a^n-b^n}{a-b}$.
Consider $\tfrac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{a-b}$. We have $f\mid a-b$ and $f\mid\tfrac{a^n-b^n}{a-b}$ so it is also the case that $f\mid\tfrac{a^n-b^n}{a-b}\pmod{a-b}$. It is relatively straightforward to show that $\sum_{i=0}^{n-1}a^{n-1-i}b^i\equiv na^{n-1}\equiv nb^{n-1}\pmod{a-b}$, therefore if $(f,n)=1$ then $f\mid a, f\mid b$. If $(f,n)=n$, then we go back to the symmetric polynomials, noting that while we haven't claimed $f^2\mid a-b$, we still have $\forall j\in[1,n], f^2\mid (p_j)$. In particular, we have $f^2\mid (p_{n-1})[u_0]$ and also $f^2\mid (p_{n-1})-(p_{n-1})[u_0]$ and $f\mid u_0$ and therefore $f^3\mid (p_n)$. But now we don't know whether $f^2\mid a-b$, and so we cannot yet extend this. We do know that $f^2\mid (a-b)^2$, so now consider
$$\sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{(a-b)^2}$$
With this particular modular equation, we cannot resolve it down to just one term on one side or the other immediately; instead, we have to systematically remove multiples of $a^2-2ab+b^2$ from both ends and meet in the middle. The first subtraction results in
$$3a^{n-2}b+\sum_{i=3}^{n-4}a^{n-1-i}b^i+3ab^{n-2}\equiv \sum_{i=0}^{n-1}a^{n-1-i}b^i\pmod{(a-b)^2}$$
while the second and further ones are similar,
$$\equiv 6a^{n-3}b^2-2a^{n-4}b^3+\sum_{i=4}^{n-5}a^{n-1-i}b^i-2a^{n-4}b^3+6a^{n-3}b^2\\ \equiv 10a^{n-4}b^3-5a^{n-5}b^4+\sum_{i=5}^{n-6}a^{n-1-i}b^i-5a^{n-5}b^4+10a^{n-4}b^3\\ \vdots\\ \equiv {m+2\choose 2}a^{m+1}b^{m-1}-\left(2\binom {m+1}2-1\right)a^mb^m+{m+2\choose 2}a^{m-1}b^{m+1}$$
Subtracting ${m+2\choose 2}a^{m-1}b^{m-1}(a-b)^2$ from this last expression yields $(2{m+2\choose 2}-2{m+1\choose 2}+1)(a^mb^m)$, and we see that $2{m+2\choose 2}-2{m+1\choose 2}+1=2m+1=n$, so now we have that $\tfrac{a^n-b^n}{a-b}\equiv na^{\tfrac {n-1}2}b^{\tfrac{n-1}2}\pmod {a^2-2ab-b^2}$, and therefore $f^2\mid na^mb^m$ and thus $f\mid ab$ ($f$ is prime). Since $f\mid ab$, now consider $\tfrac{a^n-b^n}{a-b}\equiv a^{n-1}+b^{n-1}\pmod {ab}$, which means that we have $f\mid a^{n-1}+b^{n-1}$. If we then subtract $a^{n-2}(a-b)$ from this, we get that $f\mid b^{n-1}\implies f\mid b\implies f\mid a$.
Since we have that $a-b\mid x$ and $f\mid a-b$ and now $f\mid a$ and $f\mid b$, assume that we have cleared away all such common factors, so $|f|=1$, which means that $a-b = -1$. The other possibility is that $f=a-b=0$, but this means that $x=0$ and $y=z$ which is a trivial solution. Since $f=1=1^n$, we can now rejoin our branch assumption to the other branch, namely that $f^n\mid a-b$.
This being done, our original equation looks like $x^n+y^n=(y+C^n)^n$. Now consider this equation from the perspective of $y$:
$$(y-a)^n+y^n=(y+c)^n\implies y^n+\sum_{i=0}^{n-1}{n\choose i}\left((-1)^{i+1}a^{n-i}-c^{n-i}\right)y^i=0$$
This equation is slightly different with the negative coefficients, so we create new symmetric polynomials for it with $v_k,k\in[0,n-1]$ the set of solutions and $(q_j),j\in[1,n]$ the set of polynomials. These polynomials and solutions can be defined simply, as before, and we will use $v_0$ as an intended integer solution. Rather than writing out the polynomials, we simply note important features of the set as a whole and only write out specific polynomials as needed. The first of these important features is to note that in this case, $c+a$ is a common factor to all; note that this means $c+a\mid y$. With $(q_2),(q_4),\dots,(q_{n-1})$ we have $(q_{2j})=c^{2j}-a^{2j}$ which has a factor $c^2-a^2=(c-a)(c+a)$. With $(q_1),(q_3),\dots,(q_n)$ we have $(q_{2j-1})=-c^{2j-1}-a^{2j-1}$ which has a factor $c+a$. The other important feature is that $\forall j\in[1,n-1],n\mid (q_j)$. But now we have defined the exact same scenario as before, with $c+a=b$ in the place of $a-b=c$. Finally, it is time to consider the equation from the perspective of $z$:
$$(z-b)^n+(z-c)^n=z^n$$
But wait, we have from before that $c=b-a\mid x$, and similarly $b=a+c\mid y$, which means that $\exists r,s\in\Bbb Z:rc=z-b, sb=z-c$. Rearranging, we get $(r-1)c=z-b-c=(s-1)b$ which means either $(b,c)\gt 1$ (contradicting the assumption that $(x,y,z)=1$) or else $z-b-c=bc\implies z=bc+b+c$. Since the former leads to a contradiction, we take $z=bc+b+c$ which means that we have completely specified all solutions in two parameters:
$$(bc+c)^n+(bc+b)^n=c^n(b+1)^n+b^n(c+1)^n=(bc+b+c)^n\tag 2$$
Consider the specific terms from each side which have exponent $1$ on either the $b$ or $c$ portion of the term. In particular, we see $n(bc)b^{n-1},n(bc)c^{n-1}$ appearing equally on both sides, while $nbc^{n-1}, nb^{n-1}c$ appear only on the RHS. In all other terms on both sides, there is a minimum factor $b^2c^2$. Therefore, $b^2\mid nbc,c^2\mid nbc$. We can resolve one or the other division by (w.l.o.g.) $c=1$, since we do not want $(b,c) \gt 1$. Going back to $(2)$,
$$(b+1)^n+2^nb^n=(2b+1)^n\\ \sum_{i=1}^{n}{n\choose i}b^i=\sum_{i=1}^{n-1}{n\choose i}2^ib^i\\ b^n=\sum_{i=1}^{n-1}{n\choose i}b^i(2^i-1)$$
If we divide away $b$ from both sides, we get $b^{n-1}=nbs+n$. Now we see that $b\equiv 0\pmod n$, but if we divide away $n$ from both sides, we get $b^{n-2}t=bs+1$. As $n\ge 3$, this last equation has no possible integer solutions, therefore one of our previous steps is the culprit. In particular, dividing by $b$ must have caused this issue, which would be a problem only if $b=0$. This is a valid solution, and as no other options are available, $b=0,c=1$ must be the entire set of solutions. But this means that $xyz=0$ is the only option for solutions to our original.