Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$

After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=\pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.

Our OP SunShine has correctly deduced that $\deg p(x) = 1$, so that

$p(x) = ax + b; \tag 1$

we compute:

$p^2(x) = a^2x^2 + 2ab x + b^2; \tag 2$

$p(x + 1) = ax + a + b; \tag 3$

$xp(x + 1) = ax^2 + ax + bx; \tag 4$

$xp(x + 1) + 1 = ax^2 + ax + bx + 1; \tag 5$

$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; \tag 6$

$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; \tag 7$

$a^2 = a, \; b^2 = 1, \; 2ab = a + b; \tag 8$

$a = 0, 1; \; b = \pm 1; \tag 9$

$b = 1 \Longrightarrow 2a = a + 1 \Longrightarrow a = 1 \in \{0, 1 \}; \tag{10}$

$b = -1 \Longrightarrow -2a = a - 1 \Longrightarrow a = \dfrac{1}{3} \notin \{0, 1 \}; \tag{11}$

thus the only solution is $a = b = 1$, so that

$p(x) = x + 1. \tag{12}$

CHECK:

$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$ $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; \tag{13}$

it thus appears that the only solution is

$p(x) = x + 1. \tag{14}$