I came across such a problem that I can't solve
Prove that with probability $=1$ the equality
$$W_{t}^{4} = W_{t}^{3} - 3W_{t}^{2} +1$$ holds for infinite number of $t \ge 0$ but it doesn't hold for any $t \in \mathbb{Q}_{+}$. $W_{t}$ is a standard Wiener process.
What I did: I computed expected value of both sides and end up with $$3t^2 = -3t+1$$ which has two solutions $\notin \mathbb{Q}$, one of which is positive. It makes me think that there may exist one $t$ satisfying conditions but no more than one. Is there a mistake in my attempt? If so, what is the solution of the problem?
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set $\{t\geqslant0\mid W_t=x\}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_t\in X)=0$. By countable union, $P(\exists t\in\mathbb Q_+,\,W_t\in X)=0$, qed.