The unoriginal.

From

$\begin{aligned} x+y - 1 & = \log_2 (x+y) & \small \color{#3D99F6} \text{Since } \log_b a = x \implies a = b^x \\ \implies 2^{x+y-1} & = x + y \\ \frac {2^{x+y}}2 & = x+y \\ \frac {2^{x+y}}{x+y} & = 2 \\ \implies x+y & = \begin{cases} 1 \\ 2 \end{cases} \end{aligned}$

For $x+y=1$ , then

$\begin{aligned} \log_{x+y+2} (xy+1) & = x+y - 1 & \small \color{#3D99F6} \text{Since }x+y = 1 \\ \log_3 (xy+1) & = 0 \\ \implies xy + 1 & = 3^0 = 1 & \small \color{#3D99F6} \text{Note that }y = 1-x \\ x(1-x) & = 0 \end{aligned}$

$\implies x = \begin{cases} 0 & \implies y = 1 \\ 1 & \implies y = 0 \end{cases} \implies \color{#3D99F6} \text{2 solutions}$

For $x+y=2$ , then

$\begin{aligned} \log_{x+y+2} (xy+1) & = x+y - 1 & \small \color{#3D99F6} \text{Since }x+y = 2 \\ \log_4 (xy+1) & = 1 \\ \implies xy + 1 & = 4^1 = 4 & \small \color{#3D99F6} \text{Note that }y = 2-x \\ x(2-x) + 1 & = 4 \\ \implies x^2 - 2x + 3 & = 0 & \small \color{#D61F06} \text{No real solution.} \end{aligned}$

Therefore, the number of real solutions $n=2$ and $n+x+y+1 = 2+1+1 = \boxed 4$ .