Let $p$ be an odd prime and $F={\mathbb Z}/p{\mathbb Z}$. With $a,b,c\in F$, let $Q(x,y)=ax^2+bxy+cy^2$ where $p\nmid b^2-4ac$. I wish to prove that the number of solutions $(x,y)\in F^2$ of
$$ax^2+bxy+cy^2=u$$ is the same for all units $u\in F$.


For example, when $p=5$, the equation $x^2-2xy+3y^2=u$ has one solution when $u=0$ and 6 solutions for each of $u=1,2,3,4$.



Are the number of solutions to $ax^2+bxy+cy^2\equiv u\pmod{p}$, $(x,y)\in\{0,\dotsc,p-1\}$, the same for all units $u$?
Let $F$ be a finite field, such as $\mathbf Z/(p)$ but it could be a more general finite field (including of characteristic $2$, and not necessarily $\mathbf Z/(2)$, so I won't make a change of linear variables using division by $2$ as in another answer). If both $a = 0$ and $c = 0$ then the equation becomes $bxy = u$, so necessarily $b \not= 0$ since $b^2 - 4ac = b^2$ and you said $b^2 - 4ac \not= 0$. A solution $(x,y)$ has both coordinates not equal to $0$, with $y = u/(bx)$. Thus
the solution set for each $u \in F^\times$ is $\{(x,u/(bx)) : x \in F^\times\}$, whose size is $|F| - 1$.


If $a \not= 0$ or $c \not= 0$ then by symmetry (swapping the roles of $x$ and $y$ in the equation) we can suppose $a \not= 0$. Then
solving $ax^2 + bxy + cy^2 = u$ is the same as solving $x^2 + (b/a)xy + (c/a)y^2 = u/a$, and as $u$ runs over $F^\times$ so does $u/a$. Thus we can suppose $a = 1$: we want to count solutions of $x^2 + bxy + cy^2 = u$ in $F^2$. Set $R = F[t]/(t^2 + bt + c)$, a finite ring. The norm map ${\rm N}\_{R/F} \colon R \to F$ is multiplicative, and using the basis $\{1,t\}$ we can realize your expression for $Q(x,y)$ as a norm value: for $x, y \in F$,
$$
{\rm N}\_{R/F}(x + yt) = 
\det\begin{pmatrix}x&-cy\\y&x-by\end{pmatrix} = x^2 - bxy + cy^2 = (-x)^2 + b(-x)y + cy^2.
$$
Therefore the equation $x^2 + bxy + cy^2 = u$ is the same as ${\rm N}\_{R/F}(-x+yt) = u$ for $x, y \in F$. On units the norm map ${\rm N}\_{R/F} \colon R^\times \to F^\times$ is a group homomorphism, so as with all homomorphisms between finite groups, all values are taken on an equal number of times.
Thus it remains to show the norm map ${\rm N}\_{R/F} \colon R^\times \to F^\times$ is surjective.


Case 1: $t^2 + bt + c$ is irreducible in $F[t]$. Then $R$ is a field, so $R^\times$ is cyclic and the norm map $R^\times \to F^\times$ on the nonzero elements of finite fields is onto (if $|F| = q$ then $|R| = q^2$ and a generator of $R^\times$ is mapped to a generator of $F^\times$).


Case 2: $t^2 + bt + c$ is reducible in $F[t]$. Write it as $(t-r)(t-s)$.
Since $b^2 - 4c = (r-s)^2$, from $b^2 - 4c \not= 0$ we have
$r \not= s$. Then $R \cong F[t]/(t-r) \times F[t]/(t-s) \cong F \times F$, and in the basis $\{(1,0),(0,1)\}$, the norm mapping
has the formula ${\rm N}\_{R/F}(x,y) = xy$, which maps $R^\times = F^\times \times F^\times$ onto $F^\times$.


You definitely want $b^2 - 4ac \not= 0$, since if $b^2 - 4ac = 0$ your desired conclusion can fail. As a simple example, consider $Q(x,y) = x^2 + 2xy + y^2 = (x+y)^2$. Then the equation $Q(x,y) = u$ has solutions in $F$ if $u$ is a nonzero square in $F$ and it has no solutions in $F$ if $u$ is a nonzero nonsquare in $F$. When
$F = \mathbf Z/(p)$ for odd $p$ (or more generally $F$ is a finite field of odd characteristic), half of $F^\times$ is squares and half is nonsquares, so the equation $(x+y)^2 = u$ for $u \in F^\times$ has solutions for half the values of $u$.


