$f(x) = x^2 - 1$

there are no valid solutions.

Let $ax^2 + bx + c = 0$ with $a, b, c \neq 0$. Then $f(\sqrt{2}) = (2a+c) + b \sqrt{2}$ is a root of $f$. Because $f$ has integer coefficients, $(2a+c) - b \sqrt{2}$ is the other root of $f$, so $$ax^2 + bx + c = a(x^2 - 2(2a+c) + ((2a+c)^2 - 2b^2)).$$Thus $$b = -2a(2a+c), c = a((2a+c)^2 - 2b^2).$$Substituting in gives $$c = a((2a+c)^2 - 8a^2 (2a+c)^2) = (2a+c)^2 \cdot a(1 - 8a^2).$$Write $c = ak$ for some $k$; then $$k = a^2 (2+k)^2 (1 - 8a^2).$$This implies $k < 0$; however, $|a^2 (1 - 8a^2)| \ge 7$ and $|7(2 + k)^2| > |k|$ demonstrate a solution is impossible.