The Conjecture 1 is true. We are looking for integers $m, n$ such that for sufficiently large prime $p$ we have $$ x_{1}^2\equiv 2mn\pmod{p},\quad x_{2}^2\equiv m^2-n^2\pmod{p}, \quad x_{1}^2\equiv m^2+n^2\pmod{p}. $$ To solve the first congruence we just take $m=2v^2, n=1$, where $v$ need to be specified. With $m, n$ chosen in such way we have $x_{1}=2v$ and we left with the system of congruences $$ (I)\; x_{2}^2\equiv 4v^4-1\pmod{p}, \quad (II)\; x_{3}^2\equiv 4v^4+1\pmod{p}. $$ If we substract the first from the second congruence we get that $x_{3}^2-x_{2}^2\equiv 2\pmod{p}$. To satisfy this congruence we take $$ x_{2}=\frac{1}{2}\left(\frac{1}{t}-2t\right),\quad x_{3}=\frac{1}{2}\left(\frac{1}{t}+2t\right), $$ which comes directly from the parameterization of the affine quadric curve $a^2-b^2=2$. Using the computed values of $x_{2}, x_{3}$ our congruences (I), (II) collapse to the one congruence $$ F(t,v):=4t^4-16v^4t^2+1\equiv 0\pmod{p}. $$ We thus play with the curve $C:\;F(t,v)=0$ over the finite field $\mathbb{F}_{p}$. The genus $g=g_{p}$ of the curve $C$ (depends on $p$) and is bounded by 5. Invoking now the Hasse-Weil bound, i.e., the inequality $$ |\#C(\mathbb{F}_{p})-(p+1)|\leq 2g\sqrt{p}$$ with $g=5$, we obtain that for $p+1-10\sqrt{p}>0$, i.e., for $p>100$ our curve contains a finite point.

Remark 1. The conjecture is true for $p>7$ (this follows from direct calculation).

Remark 2. One can check that if $p\equiv 1\pmod{4}$ then we can always find effectively a (rather boring) point on $C$ of the form $(t, 0)$, where $t$ is a solution of the congruence $4(4t^4+1)\equiv ((2t+1)^2+1)((2t-1)^2+1)\equiv 0\pmod{p}$. In this case we do not need to invoke Hesse-Weil bound.

The Conjecture 1 is true. We are looking for integers $m, n$ such that for sufficiently large prime $p$ we have $$ x_{1}^2\equiv 2mn\pmod{p},\quad x_{2}^2\equiv m^2-n^2\pmod{p}, \quad x_{1}^2\equiv m^2+n^2\pmod{p}. $$ To solve the first congruence we just take $m=2v^2, n=1$, where $v$ need to be specified. With $m, n$ chosen in such way we have $x_{1}=2v$ and we left with the system of congruences $$ (I)\; x_{2}^2\equiv 4v^4-1\pmod{p}, \quad (II)\; x_{3}^2\equiv 4v^4+1\pmod{p}. $$ If we substract the first from the second congruence we get that $x_{3}^2-x_{2}^2\equiv 2\pmod{p}$. To satisfy this congruence we take $$ x_{2}=\frac{1}{2}\left(\frac{1}{t}-2t\right),\quad x_{3}=\frac{1}{2}\left(\frac{1}{t}+2t\right), $$ which comes directly from the parameterization of the affine quadric curve $a^2-b^2=2$. Using the computed values of $x_{2}, x_{3}$ our congruences (I), (II) collapse to the one congruence $$ F(t,v):=4t^4-16v^4t^2+1\equiv 0\pmod{p}. $$ We thus play with the curve $C:\;F(t,v)=0$ over the finite field $\mathbb{F}_{p}$. The genus $g=g_{p}$ of the curve $C$ (depends on $p$) and is bounded by 5. Invoking now the Hasse-Weil bound, i.e., the inequality $$ |\#C(\mathbb{F}_{p})-(p+1)|\leq 2g\sqrt{p}$$ with $g=5$, we obtain that for $p+1-10\sqrt{p}>0$, i.e., for $p>100$ our curve contains a finite point.

Remark 1. One can check that if $p\equiv 1\pmod{4}$ then we can always find effectively a (rather boring) point on $C$ of the form $(t, 0)$, where $t$ is a solution of the congruence $4(4t^4+1)\equiv ((2t+1)^2+1)((2t-1)^2+1)\equiv 0\pmod{p}$. In this case we do not need to invoke Hesse-Weil bound.