Let $P \in \mathbb{Z}[i][X]$ be such that $P(0) = 1$ and $P\mod4$ is a perfect square in $(\mathbb{Z}[i] / 4\mathbb{Z}[i])[X]$. Moreover if needed we may assume $P$ is monic. Is there some way to show that in this case, it is always possible to recover polynomials $U,V \in \mathbb{Z}[i][X]$, with $P = U^2 + 4V$, such that $V(0) = 0$ and $U(0) = 1$?
In the case of $\mathbb{Z}[X]$ the answer is affirmative, but I do not know if it is possible to adapt this to the case of the coefficient ring $\mathbb{Z}[i]$.
Edit: For the $\mathbb{Z}[X]$ case, let $P = U^2 + 4V$ and write $U = a_nX^n + ... + a_1X + a_0$, then $U^2 = \sum_{k=0}^{2n} (\sum_{j=0}^n a_{j} a_{k-j})X^{k} $, and $a_0 = \pm 1 \mod 4$, wlog (by multiplying $U$ by $(-1)$), we may suppose $a_0 = 1 \mod 4$. Write $a_0 = 4m + 1$, then $U - 4m$ has constant term $1$, and $(U-4m)^2 = U^2 = P \mod 4$. Thus setting $U' = (U-4m)$, we have $P = U'^2 \mod 4$, so that $P = U'^2 + 4V'$, and $U'(0)=1$, so that $V'(0) = 0$ necessarily.
Edit: In the case of $\mathbb{Z}[X]$, letting $u = U(0)$ and $v = V(0)$, $u^2 + 4v = 1$ only has integer solutions if $v = 0$, since $1-4v$ is a square, and is thus nonnegative. Since $v$ must be an integer greater or equal to $0$ and less than or equal to $\frac{1}{4}$, $v$ must be equal to zero. (This is not correct as pointed out in the comments)
Yes, you want to show that if $P$ is a polynomial with $P(0) = 1$ and there exists $U$ with $P \equiv U^2 \pmod{4}$, we can find such $U$ with the extra constraint $U(0) = 1$. The following argument works for both $\mathbb{Z}[x]$ and $\mathbb{Z}[i][x]$. Observe that when we have $P \equiv U^2 \pmod{4}$, we can change the value of $U$ modulo $2$ since: $$(U+2V)^2 \equiv U^2 + 4V + 4 \equiv U^2 \equiv P \pmod{4}$$ In particular, we may change the constant term of $U$ modulo $2$. It suffices to show that $U(0) \equiv 1 \pmod{2}$. This is true since $U(0)^2 \equiv P(0) \equiv 1 \pmod{4}$, implying that $4 \mid (U(0)^2-1)$, so $U(0) \equiv 1 \pmod{2}$.
Remark: To see the last implication, observe that $4$ is the power of a prime (upto units) in both $\mathbb{Z}$ and $\mathbb{Z}[i]$. Thus, writing $2^2 \mid (U(0)-1)(U(0)+1)$, we have $2$ dividing one of the factors by the Pigeonhole principle. Either gives $U(0) \equiv 1 \pmod{2}$. (You could also just check all possible values of $U(0) \bmod{2}$)