Find $\lambda\in \mathbb{N}$ such that $4x^{11+\lambda}+7y^{19+\lambda}=1,~5x^6+9y^{14}=1$ has solutions. 
Find all positive integers $\lambda$ such that the system:


$$\left \{\begin {array}{lll}
4x^{11+\lambda}+7y^{19+\lambda}=1\\
5x^6+9y^{14}=1\\
\end{array}
\right.$$
has solutions.

I tried to check if some values of $\lambda$ (eg $1,2,3$) give us solutions, but the non-linearity of the system makes me think this is not a good way to start. Any help is appreciated.
Thanks in advance.
 A: COMMENT.- All curve $\Gamma$ of equation $ax^{2m}+by^{2n}=1$ where $a,b,m,n$ are positive integers is symmetric to both axis  because if $(x,y)$ is a point of $\Gamma$ so is for the four points $(\pm x,\pm y)$. Besides $\Gamma$ has a shape like a square and is contained in the square centered at the origine and having side equal to $2$.
When $m$ and $n$ are increased "in uniform way" (like as in this problem) we can see that the curves are disjoint:
in fact, for example, in the system $$\left \{\begin {array}{lll}
4x^{12}+7y^{20}=1\\
5x^6+9y^{14}=1\\\end{array}
\right.$$ we have in the first equation $x=0\Rightarrow y=\sqrt[20]{\frac14}$ and in the second one $x=0\Rightarrow y=\sqrt[14]{\frac15}$ and $$\sqrt[14]{\frac15}\lt\sqrt[20]{\frac14}\iff0.8914....\lt0.9330...$$
Similarly we can see that making $y=0$ in both equations we have values for $x$ such that the corresponding to the first equation is less than the other.
It clearly follows that there are not solutions for $\lambda$ odd.
(For $\lambda$ even the corresponding curves are not closed but we can reasoning, I guess,in akin way).
