As this question has not been answered for a very long time, I think it is appropriate to use a strong conjecture to resolve it, namely Schinzel's hypothesis H. Let us write $P(x)=P_1(x)^{k_1}...P_r(x)^{k_r}$ where $P_i(x)\in \mathbb{Z}[x]$ are irreducible, distinct, non-constant, and have positive leading coefficient (possibly $r=1$). Let the constant coefficient of $P_i$ be $c_i$. We may assume that $c_i$ are all non-zero, since otherwise we arrive at the trivial case $x\mid P(x)$. We have
\begin{align*}
\phi(n)\mid \phi(P(n))= \phi(P_1(n)^{k_1}...P_r(n)^{k_r})
\end{align*}
for all $n\geq 1.$ Let $c=|c_1...c_r|$, and define the polynomials $Q_i(x)=\frac{P_i(cx)}{|c_i|}\in \mathbb{Z}[x]$. Then we have
\begin{align*}
\phi(cn)\mid \phi(Q_1(n)^{k_1}...Q_r(n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
for all $n\geq 1$. The advantage of this was that $Q_i$ have their constant coefficients equal to $\pm 1$. Let $D$ be large enough, in particular $D>\deg P$ (but we impose another condition later). We have in particular
\begin{align*}
\phi(cD!n)\mid \phi(Q_1(D!n)^{k_1}...Q_r(D!n)^{k_r}|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
for all $n$, and these new polynomials have no small prime divisors. Finally, we apply Schinzel's hypotheisis H to the irreducible polynomials $x,Q_1(D!x),...,Q_r(D!x)$ (by Gauss' lemma, $f(x)$ is irreducible in $\mathbb{Z}[x]$ if and only if $f(kx)$ is). Their product does not have any fixed prime divisor $q$ because then we would have
\begin{align*}
\prod_{i=1}^r Q_i(D!x)\equiv 0 \mod q
\end{align*}
for $x=1,2,...,q-1$. However, this conguence has at most $\deg P$ solutions, so $q\leq \deg P+1$, which is a contradiction by the definition of $D$ (none of the polynomials $Q_i(D!x)$ is identically zero $\pmod q$ since their constant coefficients are $\pm 1$). Hence, we have infinitely many primes $p$ such that $Q_1(D!p),...,Q_r(D!p)$ are all primes. Setting $n=p$ in the previous formula involving $n$, we obtain by multiplicativity for large enough $p$ that
\begin{align*}
p-1\mid Q_1(D!p)^{k_1-1}(Q_1(D!p)-1)...(Q_r(D!p))^{k_r-1}(Q_r(D!p)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
Since $Q_i(D!p)\equiv Q_i(D!)\pmod{p-1}$, we get
\begin{align*}
p-1\mid Q_i(D!)^{k_1-1}(Q_1(D!)-1)...Q_r(D!)^{k_r-1}(Q_r(D!)-1)\phi(|c_1|^{k_1}...|c_r|^{k_r}).
\end{align*}
However, we can choose $D$, depending only on the polynomial $P$, so that $Q_i(D!)\geq 2$ for all $i$, and then the right-hand side of the previous divisibility relation should be smaller than the left-hand side, which is a contadiction for $p$ large enough.
I doubt that this problem can be solved without assuming some conjecture as $\phi(n)$ is surprisingly hard to control for general $n$; for example, Lehmer's totient problem about solving $\phi(n)\mid n-1$ seems perhaps easier but is known to be open.
So, to gain some control, one would like to choose $n$ in the relation $\phi(n)\mid \phi(P(n))$ to be prime, or at least closely related to them. However, very little is known about prime values of polynomials, and even less about their simultaneous prime values. In fact, we do not even know (according to this paper by M.Filazeta) a polynomial $f(x)\in \mathbb{Z}[x]$ such that $\deg f\geq 4$ and $f(x)$ is square-free for infinitely many integers $x$, although most numbers are squarefree and any polynomial with no fixed prime divisor should work. There are results about polynomials having infinitely almost prime values, but I am not sure whether they would be very helpful here.
Suppose there are integers $r,s$ such that $rs=255, \ r+s=1253$.
From $rs=255$ we get that both $r$ and $s$ are odd and from $r+s=1253$ a contradiction.
Best Answer
For $P\in\mathbb{Z}[t]$ and $d\in\mathbb{Z}$, let $S(P,d)$ be the set of pairs $(x,y)$ of distinct integers such that $P(x)-P(y)=d$.
The goal is to find all pairs $(P,d)$ such that $S(P,d)$ is infinite.
I'll give a solution for the case $d\ne 0$, and a partial solution for the case $d=0$ . . .
Case $(1)$:$\;d\ne 0$.
Thus, let $d$ be a nonzero integer, and let $K$ be the set of integer divisors of $d$.
Claim:$\;S(P,d)$ is infinite if and only if $P=at+b$ for some integer $b$ and some $a\in K$.
First suppose $P=at+b$ for some integer $b$ and some $a\in K$.
Write $d=ak$, for some $k\in K$.
Then for $x,y\in\mathbb{Z}$, we have \begin{align*} &P(x)-P(y)=d \qquad\qquad\qquad\qquad\qquad\qquad \\[4pt] \iff\;&(ax+b)-(ay+b)=d\\[4pt] \iff\;&a(x-y)=d\\[4pt] \iff\;&x-y=k\\[4pt] \end{align*} hence $(y+k,y)\in S(P,d)$ for all integers $y$, so $S(P,d)$ is infinite.
Conversely, suppose $S(P,d)$ is infinite.
Then for $x,y\in\mathbb{Z}$ with $x\ne y$, \begin{align*} &P(x)-P(y)=d \qquad\qquad\qquad\qquad\qquad\qquad \\[4pt] \implies\;&(x-y){\,\mid\,}d\\[4pt] \implies\;&x=y+k,\;\text{for some}\;k\in K\\[4pt] \end{align*} Since $S(P,d)$ is infinite and $K$ is finite, it follows that there is some $k\in K$ such that the equation $$ P(y+k)-P(y)=d \qquad\qquad\qquad\qquad $$ holds for infinitely many integers $y$.
Then letting $f\in\mathbb{Z}[t]$ be given by $f(t)=P(t+k)-P(t)-d$, we get that $f$ has infinitely many zeros, hence $f$ must be identically $0$. \begin{align*} \text{Then}\;&f=0\\[4pt] \implies\;&P(t+k)-P(t)-d=0\\[4pt] \implies\;&P'(t+k)-P'(t)=0\\[4pt] &\;\;\;\text{[where $P'\in\mathbb{Z}[t]$ is the formal derivative of $P$]}\\[4pt] \implies\;&P'=a,\;\text{for some integer}\;a\\[4pt] \implies\;&P=at+b,\;\text{for some integers}\;a,b\\[4pt] \end{align*} Then letting $x,y$ be distinct integers such that $P(x)-P(y)=d$, we get \begin{align*} &P(x)-P(y)=d\\[4pt] \implies\;&(ax+b)-(ay+b)=d \qquad\qquad\qquad\qquad\qquad \\[4pt] \implies\;&a(x-y)=d\\[4pt] \implies\;&a\in K\\[4pt] \end{align*} so $P=at+b$ for some integer $b$ and some $a\in K$, as claimed.
This completes the analysis for case $(1)$.
Case $(2)$:$\;d=0$.
For this case, I'll identify a class of polynomials $f\in\mathbb{Z}[t]$ such that $S(f,0)$ is infinite, and I'll conjecture that there are no others.
Let $V$ be the set of all polynomials $f\in\mathbb{Z}[t]$ such that $S(f,0)$ is infinite.
Some properties of the set $V$ . . .
Property $(0)$:$\;$If $f\in\mathbb{Z}[t]$ and $\deg(f)$ is odd, then $f\not\in V$.
Proof:$\;$Since $\deg(f)$ is odd, there exists $\theta\in (0,\infty)$ such that if $x,y\in\mathbb{R}$ with $x\ne y$ satisfy the equation $f(x)=f(y)$, then $x,y\in (-\theta,\theta)$.
Thus $(x,y)\in S(f,0)$ implies $x,y\in (-\theta,\theta)$, hence $S(f,0)$ is finite, so $f\not\in V$.
Property $(1)$:$\;t^2\in V$.
Proof:$\;$For all nonzero integers $y$, we have $(-y,y)\in S(t^2,0)$, hence $S(t^2,0)$ is infinite, so $t^2\in V$.
Property $(2)$:$\;t^2+t\in V$.
Proof:$\;$Let $f=t^2+t$.
Let $y\in\mathbb{Z}$ and let $x=-y-1$.
Then we have $x\ne y$ and $$f(x)=x^2+x=(-y-1)^2+(-y-1)=(y^2+2y+1)-(y+1)=y^2+y=f(y)$$ thus for all $y\in\mathbb{Z}$, we have $(-y-1,y)\in S(t^2+t,0)$, hence $S(f,0)$ is infinite, so $t^2+t\in V$.
Property $(3)$:$\;$If $f\in V$, then $f(t+a)\in V$ for all $a\in\mathbb{Z}$.
Proof:$\;$Let $a\in\mathbb{Z}$ and let $g=f(t+a)$.
Then $(x,y)\in S(f,0)$ implies $(x-a,y-a)\in S(g,0)$, hence $S(g,0)$ is infinite, so $g\in V$.
Property $(4)$:$\;$If $f\in V$ and $g\in\mathbb{Z}[t]$, then $g\circ f\in V$.
Proof:$\;$Let $(x,y)\in S(f,0)$. \begin{align*} \text{Then}\;\;&(x,y)\in S(f,0) \qquad\qquad\qquad\qquad\qquad\qquad\qquad \\[4pt] \implies\;&f(x)=f(y)\\[4pt] \implies\;&g(f(x))=g(f(y))\\[4pt] \implies\;&(g\circ f)(x)=(g\circ f)(y)\\[4pt] \implies\;&(x,y)\in S(g\circ f,0)\\[4pt] \end{align*} hence $S(g\circ f,0)$ is infinite, so $g\circ f\in V$.
Combining properties $(1),(2),(3),(4)$, we get
Property $(*)$:$\;$If $f=g\bigl((t+a)^2\bigr)$ or $f=g\bigl((t+a)^2+(t+a)\bigr)$ where $g\in\mathbb{Z}[t]$ and $a\in\mathbb{Z}$, then $f\in V$.
What about the converse of property $(*)$?
Conjecture:$\;$If $f\in V$, then for some $g\in\mathbb{Z}[t]$ and some $a\in\mathbb{Z}$, either $f=g\bigl((t+a)^2\bigr)$ or $f=g\bigl((t+a)^2+(t+a)\bigr)$.