Selected Problems from the Final Round of National Olympiad

Consider the following cases.
If $w = 1$, then no solution can exist, since the l.h.s. of the equality equals $2$ while the r.h.s. equals $1$.
If $w \ge 2$, then $x < z$ and $y < z$, i.e., $x \le z - 1$ and $y \le z - 1$. Thus $w^x + w^y \le w^{z-1} + w^{z-1} = 2 \cdot w^{z-1} \le w \cdot w^{z-1} = w^z$. To satisfy the equation, equalities must hold in both inequalities and thus $x = y = z - 1$ and $w = 2$. This gives the solutions $(w, x, y, z) = (2, n, n, n+1)$, where $n$ is an arbitrary positive integer.

In the case $w = 1$ there are no solutions because $1 + 1 = 2$ is not a power of $1$. Assume in the rest that $w > 1$. W.l.o.g., assume $x \le y < z$. Then the equation takes the form $w^x(1 + w^{y-x}) = w^z$, whence $1 + w^{y-x} = w^{z-x}$. Consequently, $1 + w^{y-x}$ is a positive power of $w$ and is divisible by $w$. If $y-x$ were positive, then $w^{y-x}$ would also be divisible by $w$, whence $1$ should be divisible by $w$, which is impossible. The remaining case $y-x = 0$ leads to $2 = w^{z-x}$ that gives $w = 2$ and $z-x = 1$ as the only possibility. Hence the solutions of the equation are of the form $(w, x, y, z) = (2, n, n, n+1)$, where $n$ is any positive integer. Checking shows that all these quadruples satisfy the equation.