By Matiyasevich theorem, there is no algorithm to decide whether a given Diophantine equation $P(x_1,\dots, x_n)=0$ has a solution in positive integers. As suggested in What is the smallest unsolved Diophantine equation?, let us arrange the equations by $H(P)=\sum |a_i| 2^d_i$, where $a_i$ are coefficients on the monomials of $P$ and $d_i$ are their degrees. Then consider all equations in order of $H$, and try to decide the existence of a solution in positive integers. See Can you solve the listed smallest open Diophantine equations? for a related study for all solutions (positive or negative).

After equation Positive integers such that $(x+y)(xy-1)=z^2+1$ has been solved by Denis Shatrov, I was able to solve many other equations by similar methods, including all equations of size $H\leq 25$, and almost all equations of size $H=26$. The only remaining open are:

(a) Equation $$ (x+1)yz-y-z=x^3-2. $$ It implies that $x^3-2+z$ is divisible by $y$. Write $z=ty-x^3+2$ for integer $t$, substitute in the equation, and obtain $$ t((x+1)y-1) = (x+1)(x^3-2)+1 = x^4+x^3-2x-1. $$ So, the question is whether $x^4+x^3-2x-1$ has (for some integer $x\geq 2$) a positive divisor equal to $-1$ modulo $x+1$.

(b) Equation $x^3-xy^2+y+2z^2=0$. Update: This equation has no positive integer solutions as remarked by Denis Shatrov in a comment.

(c) Equations $$ y(x^3-z^2)=z \quad \text{and} \quad x^2y^2+x=z^3 $$ In the first equation, $z=yt$, where $t=x^3-z^2=x^3-(yt)^2$. Up to the names of the variables, this is the second equation. From the second, $x(xy^2+1)=z^3$, which is possible only if $x=u^3$ and $xy^2+1=v^3$, or $u^3y^2=v^3-1$. Integers of the form $u^3y^2$ are called powerful number, and the question reduces to the existence of positive integer $v$ such that $v^3-1$ is powerful.

(d) Equation $$ y(x^3-z^2)=x $$ We have $x=yt$ for $t=x^3-z^2=(yt)^3-z^2$, or $t(t^2y^3-1)=z^2$, hence $t=u^2$ and $(u^2)^2y^3-1=v^2$, or $u^4y^3=v^2+1$.

The question is, for each of these equations, whether it has a solution is positive integers. Equations (c) and (d) look difficult, but equation (a) looks doable.

By Matiyasevich theorem, there is no algorithm to decide whether a given Diophantine equation $P(x_1,\dots, x_n)=0$ has a solution in positive integers. As suggested in What is the smallest unsolved Diophantine equation?, let us arrange the equations by $H(P)=\sum |a_i| 2^{d_i}$, where $a_i$ are coefficients on the monomials of $P$ and $d_i$ are their degrees. Then consider all equations in order of $H$, and try to decide the existence of a solution in positive integers. See Can you solve the listed smallest open Diophantine equations? for a related study for all solutions (positive or negative).

After equation Positive integers such that $(x+y)(xy-1)=z^2+1$ has been solved by Denis Shatrov, I was able to solve many other equations by similar methods, including all equations of size $H\leq 25$, and almost all equations of size $H=26$. The only remaining open are:

(a) Equation $$ (x+1)yz-y-z=x^3-2. $$ It implies that $x^3-2+z$ is divisible by $y$. Write $z=ty-x^3+2$ for integer $t$, substitute in the equation, and obtain $$ t((x+1)y-1) = (x+1)(x^3-2)+1 = x^4+x^3-2x-1. $$ So, the question is whether $x^4+x^3-2x-1$ has (for some integer $x\geq 2$) a positive divisor equal to $-1$ modulo $x+1$.

(b) Equation $x^3-xy^2+y+2z^2=0$. Update: This equation has no positive integer solutions as remarked by Denis Shatrov in a comment.

(c) Equations $$ y(x^3-z^2)=z \quad \text{and} \quad x^2y^2+x=z^3 $$ In the first equation, $z=yt$, where $t=x^3-z^2=x^3-(yt)^2$. Up to the names of the variables, this is the second equation. From the second, $x(xy^2+1)=z^3$, which is possible only if $x=u^3$ and $xy^2+1=v^3$, or $u^3y^2=v^3-1$. Integers of the form $u^3y^2$ are called powerful number, and the question reduces to the existence of positive integer $v$ such that $v^3-1$ is powerful.

(d) Equation $$ y(x^3-z^2)=x $$ We have $x=yt$ for $t=x^3-z^2=(yt)^3-z^2$, or $t(t^2y^3-1)=z^2$, hence $t=u^2$ and $(u^2)^2y^3-1=v^2$, or $u^4y^3=v^2+1$.

The question is, for each of these equations, whether it has a solution is positive integers. Equations (c) and (d) look difficult, but equation (a) looks doable.

By Matiyasevich theorem, there is no algorithm to decide whether a given Diophantine equation $P(x_1,\dots, x_n)=0$ has a solution in positive integers. As suggested in What is the smallest unsolved Diophantine equation?, let us arrange the equations by $H(P)=\sum |a_i| 2^d_i$, where $a_i$ are coefficients on the monomials of $P$ and $d_i$ are their degrees. Then consider all equations in order of $H$, and try to decide the existence of a solution in positive integers. See Can you solve the listed smallest open Diophantine equations? for a related study for all solutions (positive or negative).

After equation Positive integers such that $(x+y)(xy-1)=z^2+1$ has been solved by Denis Shatrov, I was able to solve many other equations by similar methods, including all equations of size $H\leq 25$, and almost all equations of size $H=26$. The only remaining open are:

(a) Equation $$ (x+1)yz-y-z=x^3-2. $$ It implies that $x^3-2+z$ is divisible by $y$. Write $z=ty-x^3+2$ for integer $t$, substitute in the equation, and obtain $$ t((x+1)y-1) = (x+1)(x^3-2)+1 = x^4+x^3-2x-1. $$ So, the question is whether $x^4+x^3-2x-1$ has (for some integer $x\geq 2$) a positive divisor equal to $-1$ modulo $x+1$.

(b) Equation $$ x^3-xy^2+y+2z^2=0. $$ Multiplying by $-4x$ and rearranging, we obtain $(2xy-1)^2-8xz^2=4x^4+1$.

(c) Equations $$ y(x^3-z^2)=z \quad \text{and} \quad x^2y^2+x=z^3 $$ In the first equation, $z=yt$, where $t=x^3-z^2=x^3-(yt)^2$. Up to the names of the variables, this is the second equation. From the second, $x(xy^2+1)=z^3$, which is possible only if $x=u^3$ and $xy^2+1=v^3$, or $u^3y^2=v^3-1$. Integers of the form $u^3y^2$ are called powerful number, and the question reduces to the existence of positive integer $v$ such that $v^3-1$ is powerful.

(d) Equation $$ y(x^3-z^2)=x $$ We have $x=yt$ for $t=x^3-z^2=(yt)^3-z^2$, or $t(t^2y^3-1)=z^2$, hence $t=u^2$ and $(u^2)^2y^3-1=v^2$, or $u^4y^3=v^2+1$.

The question is, for each of these equations, whether it has a solution is positive integers. Equations (c) and (d) look difficult, but (a) and (b) look doable.

By Matiyasevich theorem, there is no algorithm to decide whether a given Diophantine equation $P(x_1,\dots, x_n)=0$ has a solution in positive integers. As suggested in What is the smallest unsolved Diophantine equation?, let us arrange the equations by $H(P)=\sum |a_i| 2^d_i$, where $a_i$ are coefficients on the monomials of $P$ and $d_i$ are their degrees. Then consider all equations in order of $H$, and try to decide the existence of a solution in positive integers. See Can you solve the listed smallest open Diophantine equations? for a related study for all solutions (positive or negative).

After equation Positive integers such that $(x+y)(xy-1)=z^2+1$ has been solved by Denis Shatrov, I was able to solve many other equations by similar methods, including all equations of size $H\leq 25$, and almost all equations of size $H=26$. The only remaining open are:

(a) Equation $$ (x+1)yz-y-z=x^3-2. $$ It implies that $x^3-2+z$ is divisible by $y$. Write $z=ty-x^3+2$ for integer $t$, substitute in the equation, and obtain $$ t((x+1)y-1) = (x+1)(x^3-2)+1 = x^4+x^3-2x-1. $$ So, the question is whether $x^4+x^3-2x-1$ has (for some integer $x\geq 2$) a positive divisor equal to $-1$ modulo $x+1$.

(b) Equation $x^3-xy^2+y+2z^2=0$. Update: This equation has no positive integer solutions as remarked by Denis Shatrov in a comment.

(c) Equations $$ y(x^3-z^2)=z \quad \text{and} \quad x^2y^2+x=z^3 $$ In the first equation, $z=yt$, where $t=x^3-z^2=x^3-(yt)^2$. Up to the names of the variables, this is the second equation. From the second, $x(xy^2+1)=z^3$, which is possible only if $x=u^3$ and $xy^2+1=v^3$, or $u^3y^2=v^3-1$. Integers of the form $u^3y^2$ are called powerful number, and the question reduces to the existence of positive integer $v$ such that $v^3-1$ is powerful.

(d) Equation $$ y(x^3-z^2)=x $$ We have $x=yt$ for $t=x^3-z^2=(yt)^3-z^2$, or $t(t^2y^3-1)=z^2$, hence $t=u^2$ and $(u^2)^2y^3-1=v^2$, or $u^4y^3=v^2+1$.

The question is, for each of these equations, whether it has a solution is positive integers. Equations (c) and (d) look difficult, but equation (a) looks doable.