Let $\rm gcd(x,y)=d,x=dx_1,y=dy_1$. $\rm x_1d+y_1^2d^2+d^2=x_1dy_1d\cdot d\Leftrightarrow x_1+y_1^2d+d=d^2x_1y_1\Rightarrow d\mid x_1$,let $\rm x_1=kd,k\in \mathbb{N^*}$. $\rm kd+y_1^2d+d=d^2kdy_1\Leftrightarrow k+y_1^2+1=d^2ky_1 \Rightarrow y_1\mid k+1$,let $\rm k+1=ay_1\Leftrightarrow k=ay_1-1,a\in \mathbb{N^*}$ $\rm ay_1-1+y_1^2+1=d^2(ay_1-1)y_1\Leftrightarrow a+y_1=d^2(ay_1-1)\,\,(1)$. If $\rm d\geq2$ then $\rm d\geq 2\Rightarrow a+y_1\geq 4(ay_1-1)\Leftrightarrow y_1+4\geq a(4y_1-1)\geq4y_1-1\Rightarrow 5\geq 3y_1\Rightarrow y_1=1,a=1\,\,(2)$ $\rm (1)\overset{(2)}\Rightarrow 2=d^2\cdot 0=0$ ,contradiction! So $\rm d=1\,\,(3)$ ,therefore $\rm x+y^2+1=xy\Leftrightarrow y^2+1=x(y-1)\Rightarrow y-1\mid y^2+1\Leftrightarrow $ $\rm \Leftrightarrow y-1\mid y^2+1-(y^2-y)=y+1\Rightarrow y-1\mid 2\Rightarrow y=2\,\,or \,\,y=3$ $\rm y=2\Rightarrow x=5,y=3\Rightarrow x=5$. So $\rm (x,y)=(5,2),(5,3)$

Let $x=ak$ and $y=bk$ with $\gcd(a,b)=1$. Then the equation becomes $a+b^2k + k = abk^2 \Leftrightarrow a = k(abk - b^2 - 1)$, Then $k$ must divide $a$. Let $a=pk$. Subbing back gives us $pk+b^2k+k=pbk^3 \Leftrightarrow p + b^2 + 1 = pbk^2 \Leftrightarrow b(b+k^2) = (p+1)(bk^2-1)$. Note that since $\gcd(b, bk^2-1) = 1 $, thus $b \mid p+1$, which implies that $bk^2-1 \mid b+k^2 \Rightarrow b+k^2 \ge bk^2 - 1 \Leftrightarrow b+1 \ge k^2(b-1)$. If $k \ge 2$, then $b+1 \ge k^2(b-1) \ge 4b-4 \Rightarrow 3b \le 5 \Rightarrow b=1$. Hence $k^2 - 1 \mid k^2 + 1$, Which has no solution, therefore $k=1 \Rightarrow b-1\mid b+1 \Rightarrow b = 2,3$. From the equation above, we get that the following results are $(x,y)=(5,2),(5,3)$ . The solution is simmilar above lol