A google search of "factor square property" led only to this present question. So recall from comment that a polynomial $f(x)$ which divides $f(x^2)$ is said to be FSP or to have the factor square property.
This addresses [1], and shows an irreducible FSP quadratic must be $ax^2+ax+a$, which depending on what "irreducible" means might also imply $a=1$. We also address [3], giving a source of examples.
Let the quadratic be $p=ax^2+bx+c$, so its value at $x^2$ is $q=ax^4+bx^2+c$. If $p$ is to be a divisor of $q$ let the other factor be $dx^2+ex+f.$ Equating coefficients gives equations
[1] $ad=a,$
[2] $ae+bd=0,$
[3] $af+be+cd=b,$
[4] $bf+ce=0,$
[5] $cf=c.$
Now we know $a,c$ are nonzero (else $p$ is not quadratic, or is reducible). So from [1] and [5] we have $d=f=1.$ Then from [2] and [4] we obtain $ae=ce.$ Here $e=0$ leads to $b=0$ from either [2] or [4], and [3] then reads $a+c=0$, so that $p=a(x^2-1)$ which is reducible. So we may assume $e$ is nonzero, and also $a=c.$
At this point, [2] and [4] say the same thing, namely $ae+b=0.$ So we may replace $b=-ae$ in [3] (with its $c$ replaced by $a$) obtaining
$a+(-ae)e+a=-ae,$ which on factoring gives $a(2-e)(e+1)=0.$ The possibility $e=2$ then leads after some algebra to $2a+b=0$ and $p=a(x-1)^2$ which is reducible, while the possibility $e=-1$ leads to $a=b$ and then $p=ax^2+ax+a$ as claimed.
For some higher degree examples, for any odd prime $p$ the cyclotomic polynomial $f(x)=(x^p-1)/(x-1)$ is seen to be FSP because
$$\frac{f(x^2)}{f(x)}=\frac{x^{2p}-1}{x^2-1}\cdot \frac{x-1}{x^p-1} \\
=\frac{x^p+1}{x+1}.$$ For $p=5$ this is the polynomial $$x^4+x^3+x^2+x+1.$$
Note that for odd prime $p$ these cyclotomic polynomials are known to be irreducible (there is a simple Eisenstein proof using a change of variables).
The basic idea is to factor the polynomial over $\mathbb F_p$ for one or more suitable primes $p$, and attempt to lift these factorizations to factorizations over $\mathbb Z$. There are bounds on the size of the primes $p$ one needs to consider, making this an effective algorithm.
Over $\mathbb F_p$ one can first do a number of preliminary reductions, to reduce the problem to one where the polynomial $f$ to be factored is square-free and all irreducible factors are of fixed degree $d$. A simple way to proceed is then the (probabilistic) method of Cantor and Zassenhaus: If one takes a random monic polynomial $g$ of degree $\le 2d-1$, then the gcd of $f$ and $g$ will be a non-trivial factor of $f$ with probability about $1/2$. So, just proceed taking random $g$'s and computing the gcd with $f$, until you have a complete factorization. Other algorithms (e.g., Berlekamp) exist for this final step.
The book A Course in Computational Algebraic Number Theory by H. Cohen treats these problems in Chapter 3.4 (Factorization of Polynomials Modulo $p$) and Chapter 3.5 (Factorization of Polynomials over $\mathbb Z$ or $\mathbb Q$). These two chapters don't require any knowledge of algebraic number theory, and are easy to skim through or read.
Best Answer
Hint: Assuming that $f(x)$ is a monic and non-constant polynomial, $f(x)\mid f(x^2)$ implies that all the complex roots of $f(x)$ are roots of $f(x^2)$ too, hence that the set of roots of $f(x)$ is closed with respect to squaring. By considering the $n$-th cyclotomic polynomial with $n$ being odd, we always have that $\Phi_n(x)$ is a divisor of $\Phi_n(x^2)$, hence there are polynomials with the FSP with arbitrarily large degree (the degree of $\Phi_n(x)$ is $\varphi(n)$).