Equation $$ y^2+7xyz+3x^3-2=0 $$ is solvable in integers. Take, for example, $$ x = 47699434725285831080938680589289, $$ $$ y = 80298610335148427555, $$ $$ z = -12143437727264755796194424115079504569776282. $$

Verification is straithforward, but here is the method this solution has been found. Assume that $(x,y,z)$ is a solution with $y$ odd. Then, obviously, $$ y^2-2 \equiv 0 \,(\text{mod}\,\,x) \quad \text{and} \quad 3x^3-2 \equiv 0 \, (\text{mod}\,\,y). $$ Then $X=\frac{y^2-2}{x}$ is an odd integer. We have $$ 0 \equiv X^3(3x^3-2) = X^3\left(3\left(\frac{y^2-2}{X}\right)^3-2\right)=3(y^2-2)^3-2X^3 \equiv -24-2X^3 \, (\text{mod}\,\,y). $$ Because $y$ is odd, we conclude that $$ y^2-2 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,y). $$ Then $Y=\frac{X^3+12}{y}$ is an odd integer. Then $$ 0 \equiv Y^2(y^2-2) = Y^2\left(\left(\frac{X^3+12}{Y}\right)^2-2\right)=(X^3+12)^2-2Y^2 \equiv 144-2Y^2 \, (\text{mod}\,\,X). $$ Because $X$ is odd, we conclude that $$ Y^2-72 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,Y). $$ Now let us just try odd integers $Y=1,3,5,\dots$, try $X$ divisors of $Y^2-72$, and check whether $X^3+12$ is divisible by $Y$. If it is, compute $y=\frac{X^3+12}{Y}$, then $x=\frac{y^2-12}{X}$. Then ratio $\frac{y^2+3x^3-2}{xy}$ must be an integer, and it is left to check whether this integer is divisible by $7$. For $Y$ as small as $30761$ this works: divisor $X=135177007$ returns the solution presented above.

Equation $$ y^2+7xyz+3x^3-2=0 $$ is solvable in integers. Take, for example, $$ x = 47699434725285831080938680589289, $$ $$ y = 80298610335148427555, $$ $$ z = -12143437727264755796194424115079504569776282. $$

Verification is straithforward, but here is the method this solution has been found. Assume that $(x,y,z)$ is a solution with $y$ odd. Then, obviously, $$ y^2-2 \equiv 0 \,(\text{mod}\,\,x) \quad \text{and} \quad 3x^3-2 \equiv 0 \, (\text{mod}\,\,y). $$ Then $X=\frac{y^2-2}{x}$ is an odd integer. We have $$ 0 \equiv X^3(3x^3-2) = X^3\left(3\left(\frac{y^2-2}{X}\right)^3-2\right)=3(y^2-2)^3-2X^3 \equiv -24-2X^3 \, (\text{mod}\,\,y). $$ Because $y$ is odd, we conclude that $$ y^2-2 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,y). $$ Then $Y=\frac{X^3+12}{y}$ is an odd integer. Then $$ 0 \equiv Y^2(y^2-2) = Y^2\left(\left(\frac{X^3+12}{Y}\right)^2-2\right)=(X^3+12)^2-2Y^2 \equiv 144-2Y^2 \, (\text{mod}\,\,X). $$ Because $X$ is odd, we conclude that $$ Y^2-72 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,Y). $$ Now let us just try odd integers $Y=1,3,5,\dots$, try $X$ divisors of $Y^2-72$, and check whether $X^3+12$ is divisible by $Y$. If it is, compute $y=\frac{X^3+12}{Y}$, then $x=\frac{y^2-2}{X}$. Then ratio $\frac{y^2+3x^3-2}{xy}$ must be an integer, and it is left to check whether this integer is divisible by $7$. For $Y$ as small as $30761$ this works: divisor $X=135177007$ returns the solution presented above.

Equation $$ y^2+7xyz+3x^3-2=0 $$ is solvable in integers. Take, for example, $$ x = 47699434725285831080938680589289, $$ $$ y = 80298610335148427555, $$ $$ z = -12143437727264755796194424115079504569776282. $$

Verification is straithforward, but here is the method this solution has been found. Assume that $(x,y,z)$ is a solution with $y$ odd. Then, obviously, $$ y^2-2 \equiv 0 \,(\text{mod}\,\,x) \quad \text{and} \quad 3x^3-2 \equiv 0 \, (\text{mod}\,\,y). $$ Then $X=\frac{y^2-2}{x}$ is an odd integer. We have $$ 0 \equiv X^3(3x^3-2) = X^3\left(3\left(\frac{y^2-2}{X}\right)^3-2\right)=3(y^2-2)^3-2X^3 \equiv -24-2X^3 \, (\text{mod}\,\,y). $$ Because $y$ is odd, we conclude that $$ y^2-2 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,y). $$ Then $Y=\frac{X^3+12}{y}$ is an odd integer. Then $$ 0 \equiv Y^2(y^2-2) = Y^2\left(\left(\frac{X^3-12}{Y}\right)^2-2\right)=(X^3-12)^2-2Y^2 \equiv 144-2Y^2 \, (\text{mod}\,\,X). $$ Because $X$ is odd, we conclude that $$ Y^2-72 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,Y). $$ Now let us just try odd integers $Y=1,3,5,\dots$, try $X$ divisors of $Y^2-72$, and check whether $X^3+12$ is divisible by $Y$. If it is, compute $y=\frac{X^3+12}{Y}$, then $x=\frac{y^2-12}{X}$. Then ratio $\frac{y^2+3x^3-2}{xy}$ must be an integer, and it is left to check whether this integer is divisible by $7$. For $Y$ as small as $30761$ this works: divisor $X=135177007$ returns the solution presented above.

Equation $$ y^2+7xyz+3x^3-2=0 $$ is solvable in integers. Take, for example, $$ x = 47699434725285831080938680589289, $$ $$ y = 80298610335148427555, $$ $$ z = -12143437727264755796194424115079504569776282. $$

Verification is straithforward, but here is the method this solution has been found. Assume that $(x,y,z)$ is a solution with $y$ odd. Then, obviously, $$ y^2-2 \equiv 0 \,(\text{mod}\,\,x) \quad \text{and} \quad 3x^3-2 \equiv 0 \, (\text{mod}\,\,y). $$ Then $X=\frac{y^2-2}{x}$ is an odd integer. We have $$ 0 \equiv X^3(3x^3-2) = X^3\left(3\left(\frac{y^2-2}{X}\right)^3-2\right)=3(y^2-2)^3-2X^3 \equiv -24-2X^3 \, (\text{mod}\,\,y). $$ Because $y$ is odd, we conclude that $$ y^2-2 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,y). $$ Then $Y=\frac{X^3+12}{y}$ is an odd integer. Then $$ 0 \equiv Y^2(y^2-2) = Y^2\left(\left(\frac{X^3+12}{Y}\right)^2-2\right)=(X^3+12)^2-2Y^2 \equiv 144-2Y^2 \, (\text{mod}\,\,X). $$ Because $X$ is odd, we conclude that $$ Y^2-72 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,Y). $$ Now let us just try odd integers $Y=1,3,5,\dots$, try $X$ divisors of $Y^2-72$, and check whether $X^3+12$ is divisible by $Y$. If it is, compute $y=\frac{X^3+12}{Y}$, then $x=\frac{y^2-12}{X}$. Then ratio $\frac{y^2+3x^3-2}{xy}$ must be an integer, and it is left to check whether this integer is divisible by $7$. For $Y$ as small as $30761$ this works: divisor $X=135177007$ returns the solution presented above.