Every solution of the given system
$\left\{ \begin{array}{c} x^{2}+y=7 \\ y^{2}+x=11 \end{array} \right. \tag{0}$
is a solution of
$\left\{ \begin{array}{c} \left( y-3\right) \left( y^{3}+3y^{2}-13y-38\right) =0 \\ x^{2}=121-22y^{2}+y^{4}. \end{array}\tag{1} \right. $
The same applies to the system
$\left\{ \begin{array}{c} y^{2}=49-14x^{2}+x^{4} \\ \left( x-2\right) \left( x^{3}+2x^{2}-10x-19\right) =0. \end{array}\tag{2} \right. $
The integral solution of $(0)$ is $\left( x_{0},y_{0}\right) =\left( 2,3\right) $. Simple ways to find the remaining solutions are only possible in particular cases, as far as I know. The standard way to solve a cubic equation such as
$y^{3}+3y^{2}-13y-38=0\tag{3}$
is to make the change of variables $y=s-\dfrac{3}{3\cdot 1}=s-1\tag{3a}$ to get the reduced cubic equation
$s^{3}-16s-23=0.\tag{4}$
If the discriminant $q^{2}+\frac{4p^{3}}{27}$ of an equation of the form $s^3+px+q=0$ is negative, its three solutions are real numbers. In this case we have $q^{2}+\frac{4p^{3}}{27}=23^{2}-\frac{4\times 16^{3}}{27}<0$ and the solutions of $(4)$ can be written in the trigonometric form$^{1}$
$s_{k}=2\sqrt{\frac{16}{3}}\cos \left( \frac{1}{3}\arccos \left( \frac{23}{2}\sqrt{\frac{27}{16^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right),\tag{5}$
with $k=1,2,3$. So $y_{k}=s_{k}-1\tag{6}$ and
$x_{k}=11-y_{k}^2.\tag{7}$
For $k=1$, we get $\left( x_{1},y_{1}\right) \approx \left( -1.8479,3.5844\right) $. And similarly for $k=2$ and $k=3$.
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$^{1}$A deduction can be found in this Portuguese post of mine.
Added. If $\Delta =q^{2}+\frac{4p^{3}}{27}<0$ the three real solutions of the following reduced cubic equation
$t^{3}+pt+q=0\tag{A}$
are given by
$t_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\left( k-1\right) \pi }{3}\right) \tag{B},$
with $k=1,2,3$.
PS. I do not find trigonometric functions nor radicals ugly. But this is just an opinion.