Find all posible $x$ given the congruence $(x+y)^p-y^p \equiv 1 \pmod{q}$ for both $pA few examples. 
$$\begin{array}{c|c|l} 
\text{prime } p & \text{prime } q & \text{possible values } x \\ \hline
3 & 5 & [1, 2, 4] \\
5 & 13 & [1, 3, 5, 8, 9, 10] \\
13 & 17 & [1, 4, 7, 10, 14, 15, 16] \\
23 & 37 & [1, 2, 3, 6, 11, 12, 15, 18, 19, 22, 29, 30, 35] \\
\end{array}$$ 
We notice that the number of solutions is limited. $x=1$ is always a solution if we take $y=q-1$. 
Is there a general way of finding possible solutions for $x$ ?

If  $z = x+y$, you are solving $z^p \equiv 1 + y^p \mod q$.  
If $\gcd(p, q-1) = 1$, the map $t \mapsto t^p$ is one-to-one on $\mathbb Z/q \mathbb Z$, so there is always one $z$.  If $\gcd(p, q-1) = g > 1$, the map is neither one-to-one nor onto, so for some $y$ there is no solution and for others there are several.  The nonzero $p$'th powers mod $q$ form a subgroup of
the multiplicative group mod $q$.   
For example, if $p = 5$ and $q = 11$, the $5$'th powers mod $11$ are 
$0, 1, 10$.  $z^5 \equiv y^5 + 1$ has solutions $y^5 = 0, z^5 = 1$ corresponding to $y=0, z = 1,3,4,5,9$, and $y^5=10, z^5 = 0$ corresponding to $z=0, y=2,6,7,8,10$.
