Determine all positive integers $x, y$ and $z$ such that

$$
x^{2}+y^{2}-15=2^{z}
$$
Since a square is congruent to 0 or 1 modulo 4, we have

$$
\begin{array}{rrr}
x^{2} & \equiv 0 \text { or } 1 & \bmod 4 \\
y^{2} & \equiv 0 \text { or } 1 & \bmod 4 \\
-15 & \equiv 1 \text { or } 1 & \bmod 4
\end{array}
$$

and consequently

$$
x^{2}+y^{2}-15 \equiv 1,2 \text { or } 3 \quad \bmod 4
$$

If $z \geqslant 2$, we have $2^{z} \equiv 0 \bmod 4$.

Therefore, there is no solution if $z \geqslant 2$.

It suffices to consider the cases $z=0$ and $z=1$.

- If $z=0$, we obtain the equation

$$
x^{2}+y^{2}=16
$$

Since a square is always positive, we have in particular

$$
x^{2} \leqslant 16
$$

and by the positivity of $x$, it suffices to consider the cases $0 \leqslant x \leqslant 4$.

- For $x=0$, the equation

$$
0^{2}+y^{2}=16
$$

admits $y=4$ as a solution.

- For $x=1$, the equation

$$
1^{2}+y^{2}=16
$$

does not admit an integer solution.

- For $x=2$, the equation

$$
2^{2}+y^{2}=16
$$

does not admit an integer solution.

- For $x=3$, the equation

$$
3^{2}+y^{2}=16
$$

does not admit an integer solution.

- For $x=4$, the equation

$$
4^{2}+y^{2}=16
$$

admits $y=0$ as a solution.

- If $z=1$, we obtain the equation

$$
x^{2}+y^{2}=19
$$

Since a square is always positive, we have in particular

$$
x^{2} \leqslant 19<25
$$

and by the positivity of $x$, it suffices to consider the cases $0 \leqslant x \leqslant 4$.

- For $x=0$, the equation

$$
0^{2}+y^{2}=17
$$

does not admit a solution.

- For $x=1$, the equation

$$
1^{2}+y^{2}=17
$$

admits $y=4$ as a solution.

- For $x=2$, the equation

$$
2^{2}+y^{2}=17
$$

does not admit an integer solution.

- For $x=3$, the equation

$$
3^{2}+y^{2}=17
$$

does not admit an integer solution.

- For $x=4$, the equation

$$
4^{2}+y^{2}=17
$$

admits $y=1$ as a solution.

Therefore, the set of solutions is

$$
\mid\{(0,4,0),(4,0,0),(4,1,1),(1,4,1)\} .
$$

Remark: It would have been possible to reduce the number of cases to check by noting that, by symmetry of the problem, we can assume $x \leqslant y$ by swapping $x$ and $y$ if necessary.