$\boxed{(x,y,z)=(1,1,4)}$ If $(x,y)\ne (1,1)$,then $(x+2y+1)^2<(x+2y)^2+2x+5y+9<(x+2y+2)^2$.Contradiction.

If $y \ge 3$ $(x+2y)^2 < (y+z)^2 < (x+2y+2)^2$ So, $y+z = x + 2y + 1$ $z = x + y +1$ Contradiction If $y <3$ $y = 1$: $(x,y,z) = (1,1,4)$ $y = 2$: $(z+2)^2 - (x+5)^2 = 10 \equiv 2 \pmod 4$ contradiction $y = 3$: $(z+3)^2 - (x+7)^2 = 26 \equiv 2 \pmod 4$ contradiction

$(x+2y+1)^2<(x+2y)^2+2x+5y+9<(x+2y+3)^2$ So, $(x+2y)^2+2x+5y+9=(x+2y+2)^2$ $2x+3y=5, x=y=1, z=4$

If, $y\ge 3$, we have, that $(x+2y)^2<(y+z)^2<(x^2+2y+2)$, hence, we have $(y+z)^2=(x+2y+1)^2$, so $y+z=x+2y+1$, so $z=x+y+1,$ which yields, a contradiction. If, $y=3$, we have no solutions, same for $y=2,$ but if $y=1,$ we have that $(x+2)^2+2x+14=(z+1)^2$, so $x^2+6x+18=z^2+2z+1$, hence, $x,z=1,4$ hence, the only triple is $\boxed{(1,1,4)}.$ Note, that this solution, has some details left out, so feel free to ask me

sman96 wrote: Solve the equation for the positive integers: $$(x+2y)^2+2x+5y+9=(y+z)^2$$ We have that: $$(x+2y+1)^2<(x+2y)^2+2x+5y+9\le(x+2y+2)^2$$with equality on the right only if $2x+3y=5$, i.e. $x=y=1$. This produces the solution $(x,y,z)=(1,1,4)$ which fits.