Obviously $x-y$ is even, so $x-y\geq 2$. $x^{2n+1}-y^{2n+1}=(x-y)(x^{2n}+\cdots +y^{2n})>2x^{2n}$ and $xyz+2^{2n+1}\leq x(x-2)\cdot 5\cdot 2^{2n}+2^{2n+1}=(5x^2-10x+2)2^{2n}$ so $5x^2-10x+2>\2\left ( \frac{x}{2} \right )^{2n}\geq 2\left ( \frac{x}{2} \right )^4$ $\Rightarrow \: (x-5)((x-5)(x^2+10x+35)+180)\leq -9\: \Rightarrow \: x<5.$ so $(x,\, y)$ is $(4,\, 2)$ or $(3,\, 1)$. $(1)$ $x=4, \, y=2$ : $4^{2n+1}-2^{2n+1}=8z+2^{2n+1}\leq 40\cdot 2^{2n}+2^{2n+1}=21\cdot 2^{2n+1}$ so $2^{2n+1}-1\leq 21$, contradiction. $(2)$ $x=3, \, y=1$ : $3^{2n+1}-2^{2n}<3^{2n+1}-1=3z+2^{2n+1}\leq 17\cdot 2^{2n}$ so $\left ( \frac{3}{2} \right )^{2n}< 6<\left ( \frac{3}{2} \right )^6$, $n=2$. Then $3^5-1=3z+2^5\: \Rightarrow \: z=70.$ $\therefore (x,\, y,\, z,\, n)=(3,\, 1,\, 70,\, 2).$