Simple Method of Solution of $X^p+Y^p=(X+1)^p$

EDIT This proof is incorrect. The calculations are right, I think, but then I jump to a completely unjustified (and untrue) conclusion at the end. My inclination is to delete this answer, but I'm leaving it at the request of the OP.

I think I know how to do this. I'll just give you my idea, and let you fill in the details.

As you said, $Y \equiv 1 \pmod p,$ and as I remarked in my comments, we know $X \equiv 0 \pmod p.$ Write $X=px, Y=py+1,$ and expand the equation by the binomial theorem.$$\begin{align}p^px^p + \sum_{k=0}^p{\binom{p}{k}p^ky^k}&=\sum_{k=0}^p{\binom{p}{k}p^kx^k}\\ \sum_{k=1}^p{\binom{p}{k}p^ky^k}&=\sum_{k=1}^{p-1}{\binom{p}{k}p^kx^k}\end{align}$$ Here we have subtracted $p^px^p$ from the sum on the right, subtracted the $k = 0$ term from both sums.(Many thanks to user159517 who pointed out the need for this step.)

Now I claim that $p^3$ divides the coefficient of each of these terms, except the $k=1$ terms. When $k=1$ the coefficient is $p^2.$ When $k\ge 3$ the coefficient is obviously divisible by $p^3.$ When $k=2,$ the coefficient is $$\binom{p}{2}p^2 = \frac{p(p-1)p^2}{2} \equiv 0 \pmod {p^3},$$ since $p$ is an odd prime.

Now dividing through by $p^2$ and reducing both sides mod $p$ gives $x \equiv y \pmod p$ since all term with $k>1$ vanish. Now we can say $Y \equiv 1 \pmod {p^2}\text{ and }X \equiv 0 \pmod {p^2}.$ (No we can't! This is FALSE even in the case $x=y$.) It should be possible to prove by induction that $Y \equiv 1 \pmod {p^n}\text{ and }X \equiv 0 \pmod {p^n}$ for every $n,$ thus proving $X=0, Y=1.$

EDIT Example for $p=3,$ requested by OP.

In this case, we have $X=3x, Y=3y+1,$ so our equation becomes $$\begin{align}(3x)^3+(3y+1)^3 &= (3x+1)^3\\27x^3+27y^3+27y^2+9y+1 &=27x^3+27x^2+9x+1\\27y^3+27y^2+9y&=27x^2+9x\\3y^3+3y^2+y&=3x^2+x\\y &\equiv x \pmod 3\end{align}$$

This is an open question. See Fermat’s Last Theorem for Amateurs by Paulo Ribenboim. Specifically following results of Catalan are shown:

For $p$ odd prime number, $0<x<y$ integers such that $x^p+y^p=(y+1)^p$, the following is true:

1. $py(y+1)$ divides $x^p-1$
2. $p\nmid x$, $p\mid x-1$
3. If $q$ is a prime dividing $y+1-x$ then $q$ divides $x-1$
4. $\gcd(x+y,y+1-x)=1$
5. $\gcd(2x-1,2y+1)=1$
6. $x$ is the only integer such that $(py^{p-1})^{1/p} < x < (p(y+1)^{p-1})^{1/p}$