18. 4. 17 $\star \star$ Find all positive integer solutions $n, x, y, z, t$ of the equation $n^{x} \mid n^{y}+n^{z}=n^{t}$.
Assume that the positive integer $n$ and $x \leqslant y \leqslant z$ satisfy the equation
$$
n^{x}+n^{y}+n^{z}=n^{t} \text {. }
$$

Obviously, $n \neq 1$. If $n=2$, then from (1) we get $1+2^{y-x}+2^{z-x}=2^{t-x}$. Clearly, it cannot be that $y>x$, so $y=x, 2+2^{z-x}=2^{t-x}$. Thus, $z-x=1, t-x=2$. Therefore, when $n=2$, we have $y=x, z=x+1, t=x+2$. It is also easy to verify that for all positive integers $x$, $2^{x}+2^{x}+2^{x+1}=2^{x+2}$.

Now assume $n \geqslant 3$. From (1) we get $1+n^{y-x}+n^{z-x}=n^{t-x}$. Since $n>2, n \nmid 1, n \nmid 2$, it must be that $y-x$ and $z=x$. Thus, $3=n^{t-x}$, which means $n=3, t-x=1$. Therefore, when $n>2$, we have $n=3, x=y=z, t=x+1$. It is easy to verify that for each positive integer $x$, $3^{x}+3^{x}+3^{x}=3^{x+1}$.

Therefore, all positive integer solutions $n, x, y, z, t$ of equation (1) that satisfy $x \leqslant y \leqslant z$ are $n=2, y=x, z=x+1, t=x+2$, and $n=3, y=x, z=x, t=x+1$, where $x$ is any positive integer.