Solution:

We show that $x$ is irrational. Suppose that $x$ is rational. Then the sequence $\left\langle a_{n}\right\rangle_{n=1}^{\infty}$ is periodic after some stage; there exist natural numbers $k, l$ such that $a_{n}=a_{n+l}$ for all $n \geq k$. Choose $m$ such that $m l \geq k$ and $m l$ is a perfect square. Let

$$m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}}, \quad l=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \ldots p_{r}^{\beta_{r}}$$

be the prime decompositions of $m, l$ so that $\alpha_{j}+\beta_{j}$ is even for $1 \leq j \leq r$. Now take a prime $p$ different from $p_{1}, p_{2}, \ldots, p_{r}$. Consider $m l$ and $p m l$. Since $p m l-m l$ is divisible by $l$, we have $a_{p m l}=a_{m l}$. Hence $d(p m l)$ and $d(m l)$ have same parity. But $d(p m l)=2 d(m l)$, since $\operatorname{gcd}(p, m l)=1$ and $p$ is a prime. Since $m l$ is a square, $d(m l)$ is odd. It follows that $d(p m l)$ is even and hence $a_{p m l} \neq a_{m l}$. This contradiction implies that $x$ is irrational.

Alternative Solution:

As earlier, assume that $x$ is rational and choose natural numbers $k, l$ such that $a_{n}=a_{n+l}$ for all $n \geq k$. Consider the numbers $a_{m+1}, a_{m+2}, \ldots, a_{m+l}$, where $m \geq k$ is any number. This must contain at least one $0$. Otherwise $a_{n}=1$ for all $n \geq k$. But $a_{r}=0$ if and only if $r$ is a square. Hence it follows that there are no squares for $n>k$, which is absurd. Thus every $l$ consecutive terms of the sequence $\left\langle a_{n}\right\rangle$ must contain a $0$ after certain stage. Let $t=\max \{k, l\}$, and consider $t^{2}$ and $(t+1)^{2}$. Since there are no squares between $t^{2}$ and $(t+1)^{2}$, we conclude that $a_{t^{2}+j}=1$ for $1 \leq j \leq 2 t$. But then, we have $2 t(>l)$ consecutive terms of the sequence $\left\langle a_{n}\right\rangle$ which miss $0$, contradicting our earlier observation.