When given an equation like $x^2+y=z^2$ and asked to solve for integer values, a common technique is to bound the left-hand side (LHS) using inequalities of the form $(x+i)^2 \leq LHS \leq (x+j)^2$, where $i$ and $j$ are integers. This approach helps to restrict the possible values of $x,y$ and $z$, making it easier to identify integer solutions by narrowing the search space.

$$ (x+2y)^2+2x+5y+9-(x+2y+1)^2 = y+8$$

If $x,y,z$ are positive, then $(x+2y)^2+2x+5y+9 > (x+2y+1)^2$.

$$(x+2y+2)^2-{(x+2y)^2+2x+5y+9} = -5 + 2x + 3y \geq  0$$

$$\implies (x+2y+2)^2 \geq {(x+2y)^2+2x+5y+9} > (x+2y+1)^2$$

So, $(x+2y+2)^2=(x+2y)^2+2x+5y+9$ or $2x+3y-5=0$. To make this condition true, $x=1,y=1$. By substituting $x=1,y=1$, we obtain $z=4$ and we are done.