Showing the polynomial $x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q}[x]$.

it does not imply that if f does not have a root then f is irreducible. Is this correct?

Yes, that is correct. For polynomials of degree $> 3$, the absence of roots in $\mathbb{Q}$ is only a necessary, but not a sufficient condition for irreducibility.

Aside from trying a substitution $x \mapsto x - a$ to achieve a form where Eisenstein's criterion is applicable, you can consider the corresponding polynomial over $\mathbb{Z}/(p)$ for a prime $p$. If $f$ is reducible, so is its image $\overline{f} \in \left(\mathbb{Z}/(p)\right)[X]$, if it has the same degree as $f$ (since $f$ here is monic, that is the case). So if $\overline{f}$ is irreducible in any $\left(\mathbb{Z}/(p)\right)[X]$, then $f$ itself is irreducible.

Note: if $\overline{f}\in \left(\mathbb{Z}/(p)\right)[X]$ is reducible, that does not imply that $f$ is reducible.

If we look at the example over $\mathbb{Z}/(2)$, we have $\overline{f}(x) = x^4+x^3+1$. It is readily seen that that has no zero in $\mathbb{Z}/(2)$, so if it were reducible, both factors would have to be quadratic. Since the constant term is $1$, a factorisation would have the form $$x^4+x^3+1 = (x^2+ax+1)(x^2+bx+1).$$ But then the coefficients of $x^3$ and $x^1$ would be equal ($a+b$). So $\overline{f}$ is irreducible.

Another way, instead of trying to apply Eisenstein's, is to assume that you can indeed factor it $(x^2+bx+c)(x^2+dx+e)=x^4+x^3+4x+1$ and arrive at a contradiction. Gauss' lemma makes your work a lot easier too.

Apply Eisenstein's criterion for $p=3$ to $$f(x-1)=x^4-3\,x^3+3\,x^2+3\,x-3$$

If $f$ can be factored, both factors must have degree $2$ becuase there is no integer root. Because of $$f(-11)=13267,f(-3)=43, f(-2)=1, f(-1)=-3, f(0)=1, f(1)=7,f(4)=337$$ one of these degree $2$-factors is $1$ at three x-values (or -1). So it is identical to $1$ (or $-1$). So $f$ is irreducible. See this link ($13267,43,3,7,337$ are primes)

But it is sufficient that $$f(-11)=13267$$ is a prime because $11 \gt 2+\max\{\frac{|a_i|}{a_4} \mid i =0,1,2,3\}$ See this and this