No! There are four values of $\lambda$ such that $x^2+4x-\lambda+2\equiv 0\pmod 7$ has a solution $x\in\mathbb{Z}$.
Note that $x^2+4x-\lambda+2=(x+2)^2-(\lambda+2)$, so you want $\lambda+2$ to be a square, i.e., $\lambda+2\equiv 0,1,2,4\pmod{7}$.
Hint $ $ Let $\,f(x) = x^2-1\,$ below and exploit the highlighted $\rm\color{#c00}{multiplicativity}$ of number of roots.
Remark $\ $ If $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ & mod $\,n.\,$ By CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn,\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{mn}&\overset{\rm \large CRT\!\!}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\
&\iff& \begin{array}{}x\equiv r_1,\ldots,r_{\color{}{\large k}}\pmod{\! m},\phantom{I^{I^{I^I}}}\ \ \color{#c00}k\ \ \rm roots\\x\equiv s_1,\ldots,s_{\large\color{}{\ell}}\pmod n,\ \ \ \ \ \ \ \ \ \,\color{#c00}{\ell}\ \ \rm roots\end{array}\\[.3em]
&\iff& \left\{ \begin{array}{}x\equiv r_{\large i}\pmod{\! m}\\x\equiv s_{\large j}\pmod n\end{array} \right\}\ \ \ \ {\rm for}\ \ \begin{array}{}1 \le i \le k\\ 1\le j\le\ell\end{array}\\
&\overset{\rm\large CRT\!}\iff& \left\{ x\equiv t_{\,\large i\, j}\!\!\!\!\pmod{\!mn} \right\}\,\ \ \underbrace{{\rm for}\ \ 1 \le i\le k,\,\ 1\le j\le\ell}_{\Large \color{#c00}{k\,\cdot\, \ell}\ \,\rm roots\ \ \ \ \ \ \ \ \ \ }\\
\end{eqnarray}\qquad\qquad$$
Best Answer
alright, there are $p^2$ ordered pairs. The only pair giving $x^2 + y^2 = 0$ is $(0,0)$ because $-1$ is not a square.
For each of $(p-1)/2$ residues, there are $p+1$ pairs, total is $ \frac{p^2 - 1}{2}$
Thus $0$ and residues total to $ \frac{p^2 + 1}{2}$
Finally, nonresidues total to $ \frac{p^2 - 1}{2},$ and there are $(p-1)/2$ non-residues. The count for each non-residue is also $p+1$
P.S. the reason the counts are all the same for nonresidues: let $n_1, n_2$ be nonresidues, so there is some nonzero $a$ with $n_2 = n_1 a^2.$ Then we have a bijection of $(x,y)$ pairs: $x^2 + y^2 = n_1 $ maps to $(ax)^2 + ( ay)^2 = n_1 a^2 = n_2$