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Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions (indeed, let $u\in\mathbb{F}_p^\times$ be arbitrary and put $v:=i/u$). If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

For $p\equiv 3\pmod{3}$, there are precisely $2$ solutions of $uv=i$ with $u=\pm v$ (because precisely one of $i$ and $-i$ is a quadratic residue), hence in this case $\#(M\cap M_i)$ always equals $\frac{p-3}{4}$.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.

Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions (indeed, let $u\in\mathbb{F}_p^\times$ be arbitrary and put $v:=i/u$). If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

For $p\equiv 3\pmod{3}$, there are precisely $2$ solutions of $uv=i$ with $u=\pm v$ (because precisely one of $i$ and $-i$ is a quadratic residue), hence in this case $\#(M\cap M_i)$ always equals $\frac{p-3}{4}$.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.

Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions (indeed, let $u\in\mathbb{F}_p^\times$ be arbitrary and put $v:=i/u$). If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.

Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions (indeed, let $u\in\mathbb{F}_p^\times$ be arbitrary and put $v:=i/u$). If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

For $p\equiv 3\pmod{3}$, there are precisely $2$ solutions of $uv=i$ with $u=\pm v$ (because precisely one of $i$ and $-i$ is a quadratic residue), hence in this case $\#(M\cap M_i)$ always equals $\frac{p-3}{4}$.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.

Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions. If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.

Here is an elementary approach. $\#(M\cap M_i)$ equals the number of solutions $y-x=i$ with $x,y\in M$. Every quadratic residue is the square of two nonzero residues, so $\#(M\cap M_i)$ equals $1/4$ times the number of solutions of $a^2-b^2=i$ with $a,b\in\mathbb{F}_p^\times$. For $p>2$ we can make the bijective change of variables $u:=a+b$ and $v:=a-b$, which shows that we are counting the number of solutions $uv=i$ with $u,v\in\mathbb{F}_p$ such that $u\neq\pm v$. Without the restriction $u\neq\pm v$, there are exactly $p-1$ solutions (indeed, let $u\in\mathbb{F}_p^\times$ be arbitrary and put $v:=i/u$). If $i$ is a quadratic residue, then there are precisely $4$ solutions with $u=\pm v$, otherwise there are no such solutions.

To summarize, $\#(M\cap M_i)$ equals $\frac{p-5}{4}$ or $\frac{p-1}{4}$ depending on whether $i$ is a quadratic residue or not.

Added. In general, it is easy to count the number of solutions of $ a_1x_1^2+\dots +a_kx_k^2=b $ in a finite field. For a treatment of $k=2$ (which is straightforward to extend to $k>2$) see my earlier response here.