Let $f\colon \Bbb Z\to \Bbb Z$ be a any function function with $$\tag0f(x-f(y)) = f(f(x)) - f(y) - 1$$
for all $x,y\in\Bbb Z$.
Letting $y=f(x)$ we find $$f(x-f(f(x)))=-1. $$
So for $a=-f(f(0))$ we have $f(a)=-1$. Then with $y=a$, $$\tag1f(x+1)=f(f(x)) $$
Then $(0)$ becomes
$$\tag2 f(x-f(y))=f(x+1)-f(y)-1 $$
Or with $g(x):=f(x)+1$ (and $x\leftarrow x-1$)
$$\tag3g(x-g(y))=g(x)-g(y)$$
From $(3)$ we see that the image of $g$ is a subgroup of $\Bbb Z$, hence it is either $\{0\}$ (in which case $f(x)=-1$), or $c\Bbb Z$ for some $c\ge 1$.
In that case, for $n\in\Bbb Z$ we find $y$ with $g(y)=nc$ and so have $g(x+ nc)=g(x)+nc$. Thus $g$ is determined by the values $g(0),\ldots, g(c-1)$. On the other hand, these values can indeed be chosen freely. In other words:
Claim 1. Let $c\in\Bbb N$ and $b_0,\ldots, b_{c-1}\in\Bbb Z$. Then the function $g$ given by
$$ g(x)= (n+b_r)c$$
where $x=nc+r$, $0\le r<c$
is a solution to $(3)$, and all non-zero solutions of $(3)$ are obtained this way.
Proof.
Let $x=nc+r$, $y=mc+s$ with $0\le r,s<c$. Then
$$\begin{align}g(x-g(y))&=g(nc+r-(m+b_s)c)\\
&=g((n-m-b_s)c+r)\\
&=(n-m-b_s+b_{r})c\\
&=(n+b_r)c-(m+b_s)c\\&=g(x)-g(y)\end{align}$$
That all non-zero solutions are of this form has been shown above. $\square$
Then the solutions $f$ of $(2)$ (apart from $f(x)=-1$) are precisely those of the form $f(x)=g(x)-1$ with $g$ as in Claim 1.
Such $f$ is a solution to the original $(0)$ if and only if we additionally have $(1)$ for all $x$.
Note that for $x=nc+r$, $0\le r<c$, we have $f(x)=g(x)-1=(n+b_r)c-1=(n+b_r-1)c+c-1$ so that $$f(f(x))=(n+b_r-1+b_{c-1})c-1.$$
On the other hand,
$$f(x+1)=g(x+1)-1=\begin{cases}(n+b_{r+1})c-1&\text{if }r<c-1\\(n+1+b_0)c-1&\text{if }r=c-1\end{cases}$$
We conclude that $b_{r+1}=b_r+b_{c-1}-1$ for $0\le r<c-1$, and that $2b_{c-1}-1=b_0+1$. From the first we see that $b_r=b_0+rb_{c-1}-r$, so
$$\begin{align}b_{c-1}&=b_0+1+(c-1)b_{c-1}-c\\
&=2b_{c-1}-1+(c-1)b_{c-1}-c\\
&=(c+1)b_{c-1}-c-1\end{align}$$ and finally
$$b_{c-1} = \frac {c+1}c.$$
This is an integer only for $c=1$ and in that case we arrive at $b_0=2$
Thus the only solutions to $(0)$ apart from $f(x)=-1$ is
$$f(x)=x+1.$$
Suppose $f:\Bbb R\to \Bbb R$ satisfies
$$f\big(f(x)-x+y^2\big)=yf(y)\tag{1}$$
for all $x,y\in \Bbb R$. Plugging in $x,y=1$ in $(1)$, we get
$$f\big(f(1)\big)=f(1).$$
Hence, substituting $x=f(1)$ in $(1)$, we obtain
$$f(y^2)=yf(y).\tag{2}$$
This shows that $f(0)=0$ and for $y\ne 0$,
$$-yf(-y)=f\big((-y)^2\big)=f(y^2)=yf(y),$$
so $f(-y)=-f(y)$. Hence $f$ is an odd function.
From $(1)$ and $(2)$, we get
$$f\big(f(x)-x+y^2\big)=f(y^2)$$
for all $x,y\in\mathbb{R}$. Substitute $-x$ for $x$ in the last equation and using the fact that $f$ is odd, we get
$$f\big(f(x)-x-y^2\big)=f(-y^2)$$
for all $x,y\in\mathbb{R}$. Hence,
$$f\big(f(x)-x+y)=f(y)\tag{3}$$
for all $x,y\in\mathbb{R}$.
Define $P$ to be the additive subgroup of $\mathbb{R}$ generated by $\big\{f(x)-x\big|x\in\Bbb R\big\}$. Then, we see that for any $z\in\mathbb{R}$ and $p\in P$, we have
$$f(z+p)=f(z).\tag{4}$$
Consequently $f$ is an odd periodic function which is invariant under translation by elements of $P$.
We claim that $P=f^{-1}(0)$. First if $p\in P$, then $f(p)=f(p+0)=f(0)=0$, so $p\in f^{-1}(0)$. Conversely, suppose that $p\in f^{-1}(0)$, then with $x=p$ in $(3)$, we have
$$f(-p+y)=f\big(f(p)-p+y\big)=f(y)$$
so that $p\in P$.
For each $p\in P$, we see that
$$f\left(\left(\frac{1+p}{2}\right)^2\right)=f\left(\left(\frac{1-p}{2}\right)^2+p\right)=f\left(\left(\frac{1-p}{2}\right)^2\right)$$
From $(2)$, we have
$$\frac{1+p}{2}f\left(\frac{1+p}{2}\right)=\frac{1-p}{2}f\left(\frac{1-p}{2}\right).$$
However
$$f\left(\frac{1+p}{2}\right)=f\left(\frac{1-p}{2}+p\right)=f\left(\frac{1-p}{2}\right).$$
This means
$$f\left(\frac{1+p}{2}\right)=0$$
for all $p\in P$ such that $p\ne 0$. Hence, for $p\in P\setminus\{0\}$, $\frac{1+p}{2}\in P$, so $1+p \in P$, making $1\in P$. This shows that either $P=\{0\}$ or $\mathbb{Z}\left[\frac12\right]\subseteq P$.
If $P=\{0\}$, then $f(x)-x=0$ for all $x\in \Bbb R$. Therefore, $f(x)=x$ for every $x\in \Bbb R$. This yields one solution of $(1)$. From now on $P\neq \{0\}$. Thus $\mathbb{Z}\left[\frac12\right]\subseteq P$. We want to show that $P=\Bbb R$, so that $f(x)=0$ for every $x\in \Bbb R$, and this is another solution of $(1)$.
For an arbitrary $y\in\mathbb{R}$, we see that
$$f\left(\left(y+\frac{1}{2}\right)^2\right)=\left(y+\frac{1}{2}\right)f\left(y+\frac{1}{2}\right)=\left(y+\frac{1}{2}\right)f(y),$$
since $1/2\in P$. That is
$$f\left(\left(y+\frac12\right)^2\right)=\frac{1}{2}f(y)+yf(y)=\frac12f(y)+f(y^2).$$
Consequently
\begin{align}f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2&=\frac{1}{2}f(y)+f(y^2)-\left(y+\frac12\right)^2\\&=\left(\frac{1}{2}f(y)-y\right)+\left(f(y^2)-y^2\right)+\frac14.\end{align}
Because $f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2$, $f(y^2)-y^2$, and $\frac14$ are all elements of $P$, we conclude that
$$\frac{1}{2}f(y)-y\in P.$$
Therefore,
$$\big(f(y)-y\big)-y=2\left(\frac{1}{2}f(y)-y\right)\in P.$$
As $f(y)-y\in P$, we get $y\in P$.
Best Answer
You can show that the functions $ f : \mathbb R \to \mathbb R $ satisfying $$ f \big( x + f ( y ) \big) = y + f ( x + 1 ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ are exactly those of the form $ f ( x ) = A ( x ) + 1 $, where $ A $ is an additive involution, i.e. $ A ( x + y ) = A ( x ) + A ( y ) $ and $ A \big( A ( x ) \big) = x $ for all $ x , y \in \mathbb R $. If further regularity conditions like continuity, local boundedness, integrability or measurability are assumed for $ f $, then $ A $ will be regular, too. The only regular additive functions are those of the form $ A ( x ) = c x $ for some constant $ c \in \mathbb R $ (see here). Thus the only regular additive involutions are $ A ( x ) = x $ and $ A ( x ) = - x $, as we must have $ A \big( A ( 1 ) \big) = c ^ 2 = 1 $. Therefore, the only regular solutions to \eqref{0} are $ f ( x ) = x + 1 $ and $ f ( x ) = - x + 1 $. Without any further conditions on $ f $, one can use the axiom of choice to show that there are nonregular solutions, too (see Example 5.6 in this PDF file).
It's straightforward to check that if $ f $ is of the form $ f ( x ) = A ( x ) + 1 $ for some additive involution $ A $, then it satisfies \eqref{0}. We try to prove the converse.
Letting $ x = 0 $ in \eqref{0} we get $$ f \big( f ( y ) \big) = y + f ( 1 ) \text . \tag 1 \label 1 $$ In particular, \eqref{1} shows that if $ f ( x ) = f ( y ) $ then $ x = y $. Letting $ y = 0 $ in \eqref{1} shows that $ f \big( f ( 0 ) \big) = f ( 1 ) $, and thus by injectivity, $ f ( 0 ) = 1 $. Hence, putting $ x = - 1 $ in \eqref{0} we have $$ f \big( f ( y ) - 1 \big) = y + 1 \text . \tag 2 \label 2 $$ \eqref{2} shows that if we substitute $ x - 1 $ for $ x $ and $ f ( y ) - 1 $ for $ y $ in \eqref{0}, we get $$ f ( x + y ) = f ( x ) + f ( y ) - 1 \text . \tag 3 \label 3 $$ Now, if we define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = f ( x ) - 1 $, then by \eqref{3} we can see that $ A $ is additive, and by \eqref{2} we can see that $ A $ is an involution. This proves what was desired.