887. Solve the equation in natural numbers

$$
2^{x}-15=y^{2}
$$
$\triangle$ Consider two cases.

1) Let $x$ be odd: $x=2 k+1(k \in N)$.

Let's see what remainder $2^{x}$ gives when divided by 3.

$$
2^{x}=2^{2 k+1}=2 \cdot 2^{2 k}=2 \cdot 4^{k}=2\left(\left(4^{k}-1\right)+1\right)=2\left(4^{k}-1\right)+2
$$

Since the difference $4^{k}-1$ is divisible by 3 (§ 7, first divisibility), the number $2^{x}$ gives a remainder of 2 when divided by 3.

$$
2^{x}=3 a+2(a \in N)
$$

Substitute this expression for $2^{x}$ into the equation

$$
3 a+2-15=y^{2}, \quad 3 a=y^{2}+13, \quad 3 a=\left(y^{2}+1\right)+12
$$

From here, the sum $y^{2}+1$ is divisible by 3. But this is impossible (see $\S 6$, statement of support problem 163$)$.

2) Let $x$ be even: $x=2 k \quad(k \in N)$.

We get:

$$
2^{2 k}-15=y^{2}, \quad 2^{2 k}-y^{2}=15, \quad\left(2^{k}+y\right)\left(2^{k}-y\right)=15
$$

Here there are two possibilities:

$$
\left\{\begin{array} { l } 
{ 2 ^ { k } + y = 1 5 , } \\
{ 2 ^ { k } - y = 1 , }
\end{array} \quad \left\{\begin{array}{l}
2^{k}+y=5 \\
2^{k}-y=3
\end{array}\right.\right.
$$

It remains to solve each of these systems of equations. Do this on your own.

Answer: $(6 ; 7),(4 ; 1)$.