When comparing models which differ by a single parameter, AIC corresponds to comparing by using an asymptotic (chi-squared) LRT on the additional parameter with $\alpha=15.73\%$.
Consider that with AIC you say the predictor should remain if AIC goes down when you include it, while with the LRT you say the predictor should remain if the chi-squared p-value is below $\alpha$.
Now $\text{AIC} = -2 \log\cal{L} + 2p$, so the AIC goes down if $-2 \log \cal{L}_0+2p_0>-2 \log \cal{L}_1+2p_1$ (where the subscript $0$ denotes the smaller model).
That is, it goes down if $-2 (\log \cal{L}_0- \log \cal{L}_1)>2(p_1-p_0)$.
Which is to say, with AIC you choose the larger model when $-2 (\log \cal{L}_0- \log \cal{L}_1)>2\times 1$. But the LHS is just the quantity used in the asymptotic chi-squared LRT, so under the null you choose the larger model when a $\chi^2_1>2$, which corresponds to using the asymptotic chi-squared test at a significance level of $0.1573$ (the tail area beyond $2$ in a $\chi^2_1$).
So your "clearer" result is nothing more than saying to the person who doubts* $p=0.04$ "well, if I use $\alpha=0.1573$, instead of $\alpha=0.05$, that $p=0.04$ looks pretty good". If they're capable of doing the calculation I just did for you, they may not be all that impressed by a desire to simply inflate a significance level because it makes the result look superficially (but not actually) "stronger".
* (perhaps somewhat inexplicably, as if p-values are all that meaningful)
[That's not to denigrate AIC in particular - indeed it may make more sense to actually use it than to use $p=0.05$ when deciding whether the additional variable is useful, but we should be clear about what it is we're choosing when we do it. If we choose it, our justification should not be based on this somewhat illusory notion of greater strength, but on its properties.]
