26. Find all integer solutions $(x, y, z)$ that satisfy the following system of equations:
$$\left\{\begin{array}{l}
x+1=8 y^{2}, \\
x^{2}+1=2 z^{2} .
\end{array}\right.$$
26. First, $(x, y, z)=(-1,0, \pm 1)$ is a solution to the system of equations. Next, consider the solutions to the system of equations when $y \neq 0$. By eliminating $x$ from the system, we get
$$1+\left(2(2 y)^{2}-1\right)^{2}=2 z^{2},$$

which simplifies to $(2 y)^{4}+\left((2 y)^{2}-1\right)^{2}=z^{2}$. Therefore, there exist $a, b \in \mathbf{N}^{*}$ such that
$$4 y^{2}=2 a b, 4 y^{2}-1=a^{2}-b^{2},(a, b)=1, a>b .$$

From the above equations, it is easy to see that $a$ is even and $b$ is odd. Thus, we can set
$$a=2 s^{2}, b=t^{2},(s, t)=1, s, t \in \mathbf{N}^{*} .$$

This leads to $4 s^{2} t^{2}-1=4 s^{4}-t^{4}$, hence $\left(2 s^{2}+t^{2}\right)^{2}=8 s^{4}+1$. Using the result from the previous problem, we know $s=1, 2 s^{2}+t^{2}=3$, which implies $s=t=1, a=2, b=1$, so $y= \pm 1, x=7, z=$ $\pm 5$.

Therefore, the integer solutions to the system of equations are $(x, y, z)=(-1,0, \pm 1),( \pm 1,7, \pm 5)$.