Do flights affect tides?
Both airborne planes and the Moon affect the tides, but the effect of airborne planes is so small as to be negligible. 


From your sources the mass of a Boeing 747 ($m_{747}$) is $447.696$
$(10)^3$ $kg$.
Likewise, from your sources, the mass of the Moon ($m_{Moon}$ is
$7.347$ $(10)^{22}$ $kg$.
Typically a 747 would fly at an altitude ($r_{747}$) of around $30$
$000$ $ft$. This is close to $10$ $000$ $m$ ($10^4$ $m$).
The distance the Moon is from the Earth ($r_{Moon}$) is $384$
$400$  $km$, which is $384.4$ $(10)^6$ $m$.


Now, Newton's Law of Universal Gravitation states, $F$ $=$ ${G.m_1.m_2}/{r^2}$

Where:
$G$ is the gravitational constant, being $6.6743$ $10^{-11}$ $m^3.kg^{-1}.s^{-2}$,
$m_1$ and $m_2$ are the masses of the two objects being considered &amp; $r$ is the distance between them.

Because tides are being considered only the mass of water is to be considered, not the mass of the Earth. To simplify things I am assuming the area of water affect by each is the same and hence the mass of water will be the same in each case.

For the Earth and Moon system, the force of gravity will be:

$F_{E-M}$ $=$ $6.6743(10)^{-11}$ $.$ $7.347(10)^{22}$ $.$ $m_{ocean}$ $/$ $[{384.4(10)^6}]^2$ $=$ $3.3186(10)^{-5}$ $.$ $m_{ocean}$

For the Earth Boeing 747 system, the force of gravity will be:

$F_{E-747}$ $=$ $6.6743(10)^{-11}$ $.$ $447.696(10)^{3}$ $.$ $m_{ocean}$ $/$ $[{(10)^4}]^2$ $=$ $2.9881(10)^{-13}$ $.$ $m_{ocean}$

Dividing $F_{E-M}$ by $F_{E-747}$ the gravitational force of the Moon is $111$ $060$ greater than that of the plane.

If the tides produced by the Moon are up to $10$ $m$, then the maximum height of a tide produced by a 747 would be around $0.099$ $mm$.