square root of $-1$ in $\mathbb{Q}_p$

Question

In Computing the quotient $\mathbb{Q}_p[x]/(x^2 + 1)$ and Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields it has been claimed that since $x^2+1$ is irreducible mod $4$ it is irreducible in $\mathbb{Q}_2[x]$ and since $x^2+1$ is irreducible mod $p$ for $p \equiv 3 \bmod 4$ it is irreducible in $\mathbb{Q}_p[x]$. I understand (by inspection for $4$ and quadratic reciprocity for $p \equiv 3 \bmod 4$) that $x^2+1$ is indeed irreducible modulo the relevant primes. But why does this imply that it is irreducible in $\mathbb{Q}_p[x]$? I have tried using Hensel's lemma, but I could only get it to show that if there is a root mod $p$, then it will lift to a root in $\mathbb{Z}_p$. This problem seems to require a converse.