If $d+1$ is not squarefree, then the multiples of it are not squarefree, therefore your value of $n$ is not squarefree, so then by the Generalized Korselt's criterion it is not a valid value of $n$, unless it equals $d+1$ itself, but that is just one case.
If $d+1$ is squarefree, we note that for every prime $p \geq d+1$ we have that $n=p(d+1)$ is squarefree. Then we want that for every prime divisior $q_i$ of $n$, that $q_i-1 \mid (p-1)(d+1)$. It surely holds for $q_i=p$. Note that since $d+1$ is squarefree, we have that $$\varphi(d+1)=(p_1-1)(p_2-1)(p_3-1) \cdots(p_t-1)$$
Here $\varphi(x)$ is the Euler-phi function.
Therefore if $\varphi(d+1) \mid p-1$, then our $n$ is a possibility. In other words, if $p = k \varphi(d+1) + 1$ for some $k$. Note that we always have $\varphi(d+1) \leq d$, so there are quite a few values.
Since $\gcd(\varphi(d+1),1)=1$, we have by a stronger version of Dirichlet's theorem that the primes in this arithmetic progression are distributed just as in the natural numbers, i.e. there are about $\frac{x}{\ln(x)}$ primes when $k<x$ ($k$ as in $p = k \varphi(d+1) + 1$). This gives some information on the second part of the conjecture.
For example, when $d=4$, $13\cdot5$, $17\cdot5$, $29 \cdot 5$ and $31 \cdot 5$, are all in your sequence, and they indeed have the form I described above.
Note: One could use $\lambda(x)$, the Carmichael function, instead of $\varphi(x)$.
The phrase "non-repeating" can be a bit confusing when first introduced. A more precise, if less snappy, term is "not eventually periodic" (and this is what mathematicians mean when they say "non-repeating" in the context in question).
A sequence of numbers $(a_i)_{i\in\mathbb{N}}$ is eventually periodic iff there are $m,k$ such that for all $n>m$ we have $a_n=a_{n+k}$. The "eventually" here is connected to the "$m$" - the sequence $$0,1,2,3,4,5,6,4,5,6,4,5,6,...$$ is not periodic but it is eventually periodic (take $m=4$ and $k=3$). On the other hand, the sequence $$0,1,0,0,1,0,0,0,1,0,0,0,0,1,...$$ is not even eventually periodic (although of course it does have lots of repetition in it).
The connection with irrationality is this:
For a real number $r$, the following are equivalent:
(The issue re: these last two bulletpoints is that a few numbers have multiple decimal expansions. But this isn't a big deal to focus on at first.) In particular, the number $$0.01001000100001000001...$$ is irrational.
And base $10$, unsurprisingly, plays no role here: the above characterization works with "decimal expansion" replaced with "base-$b$ expansion" for any $b$.
Best Answer
Let $P := \{1, x, x^2, x^3, x^4\}$. The proof relies on noticing that $$\begin{align} A(n) = x &\iff n = 0 \\ A(n) = x^2 &\iff n \in P \\ x^4 \mid A(n) &\iff n \notin P \cup \{0\} \end{align}$$
First, it is not hard to deduce the first equation, because in the recurrence we have a square. This also means that $$ \forall n \geq 1, \quad x^2 \mid A(n), $$ and more generally, $$ \forall n \geq 1, \exists k \geq 1 \quad A(n) = x^{2^k}. \tag{1} $$
For $n \in \{1,\ldots, x^4\} \setminus P$, we have $n \nmid A(n-1)$ because $A(n-1)$ is a power of $x$ and $n$ is not. So $N := A(n-1) \bmod n \geq 1$. This means, by $(1)$, there is $k\geq1$ such that $$ A(n) = A(N)^2 = (x^{2^k})^2 = x^{2^{k+1}} $$ which implies $$ \forall n \in \{1,\ldots, x^4\} \setminus P, \quad x^4 \mid A(n). \tag{2} $$ If $n \in P$, then $n-1 \notin P$. By $(2)$ This means $x^4 \mid A(n-1)$ which implies $n \mid A(n-1)$ (because $n \mid x^4$). So $$ A(n) = A(A(n-1) \bmod n)^2 = A(0)^2 = x^2. $$ In other words, $$ \forall n \in P, \quad A(n) = x^2. \tag{3} $$
Now we are able to prove the full statement by induction, that is for $k \geq 1$, $A(x^4+k) = x^4$. At $k=1$, we have by $(3)$ ($x^2 \in P$) $$ A(x^4+1) = A(A(x^4) \bmod x^4+1)^2 = A(x^2)^2 = x^4. $$ Suppose the property true for some $k \geq 1$, then by $(3)$ ($x^4 \in P$) $$ A(x^4+k+1) = A(A(x^4+k) \bmod x^4 + k +1)^2 = A(x^4)^2 = x^4. $$ The property holds for $k+1$, which complete the induction.