Let's find a natural number that is one less than twice a square number, and the same can be said about its square.
With a few trials, we can find that the numbers 1 and 7 satisfy the condition. Indeed,

$$
1=1^{2}=2 \cdot 1^{2}-1 ; \quad 7=2 \cdot 2^{2}-1, \quad 7^{2}=2 \cdot 5^{2}-1
$$

Remark. We do not know if there are any other solutions. The problem seeks such integers $x, y, z$ that satisfy the equations

$$
x=2 y^{2}-1, \quad x^{2}=2 z^{2}-1
$$

Substituting $x$ from the first equation into the second, the resulting equation can be transformed into

$$
\left(y^{2}\right)^{2}+\left(y^{2}-1\right)^{2}=z^{2}
$$

which means that $y^{2}, y^{2}-1, z$ form a Pythagorean triplet for values of $y$ and $z$ that satisfy the conditions. Using the known properties of these triplets, the search for further solutions can be reduced to finding integers $r$ and $s$ that satisfy the equation $4 r^{2} s^{2}-1=4 r^{4}-s^{4}$ or, in another form, $\left(2 r^{2}-s^{2}\right)^{2}-2\left(s^{2}\right)^{2}=-1$. (With these, $y$ and $z$ can be expressed as $y=2 r s, z=4 r^{4}+s^{4}$.) The equation has a solution $r=s=1$. (This leads to the above value $x=7$.) We do not know if the equation has any other solutions. We do know that the equation $u^{2}-2 v^{2}=-1$ has infinitely many integer solutions $(u, v)$, and there are some where $v$ is a perfect square (e.g., 239, 169, but their sum is not twice a perfect square).

Most contestants provided various incorrect proofs that there are no other solutions besides the two mentioned above. We accepted as correct (3 points) any solutions that included the two values mentioned above; there were 35 such solutions; another 37 were incomplete (1 point).