Is there an integer-sided right triangle with square perimeter and square hypotenuse? Is there an integer-sided right triangle with square perimeter and square hypotenuse?

Frenicle[89] noted (pp. 71-8) that if the hypotenuse and perimeter of a right triangle both are squares, the perimeter has at least 13 digits.
"History Of The Theory Of Numbers Vol-II" by Leonard Eugene Dickson
Chapter 4, p187
https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n213

Let's suppose the right triangle be $\left(x,y,q^2\right)$,
then perimeter is $\left(x+y+q^2\right)$
\begin{align*}
x\!^{\phantom{1}}\,+\,y\!^{\phantom{1}}&=p^2-q^2\\
x^2+y^2&=q^4
\end{align*}
(not sure) I am pessimistic that there is no such integer-sided right triangle.
 A: I considered this problem in the following paper
"TWO EXTREME DIOPHANTINE PROBLEMS CONCERNING THE PERIMETER OF PYTHAGOREAN TRIANGLES"
in the journal Glasnik Matematicki, Vol. 46, No.1 (2011), 1-5.
I showed that no such triangles exist.
A: An elementary proof for primitive triangles
A primitive triangle has sides $2mn,m^2-n^2,m^2+n^2$ where $m,n$ are coprime and of opposite parity. We require $m^2+n^2$ and $2m(m+n)$ to be squares.
Then $m=2u^2, n=v^2-2u^2$ and $$v^4-4v^2u^2+8u^4=z^2,$$ where $u,v,z$ are coprime in pairs and $v,z$ are odd.
Then $(v^2-2u^2)^2-z^2=-4u^4$ and so
$$\left(\frac{v^2-2u^2+z}{2}\right)\left(\frac{v^2-2u^2-z}{2}\right)=-u^4$$
The two bracketed expressions differ by $z$ and so have no common factor with $u$. Therefore there are coprime integers $r,s$ such that $u=rs$ and $$ \left\{\frac{v^2-2u^2+z}{2},\frac{v^2-2u^2-z}{2}\right\}=\{r^4,-s^4\}$$
Then $v^2-2u^2=r^4-s^4$ and  $r^4+2r^2s^2-s^4=v^2.$ Complete the square again.
$$(r^2+s^2)^2-v^2=2s^4$$ $$(r^2+s^2+v)(r^2+s^2-v)=2s^4$$
Similarly to before, we obtain coprime integers $X,Y$ such that $s=2XY$, with 
$$\{r^2+s^2+v,r^2+s^2-v\}=\{2X^4,16Y^4\}$$
Then $$X^4-4X^2Y^2+8Y^4=r^2.$$ 
Now $u=2rXY$ and so comparing this equation with $v^4-4v^2u^2+8u^4=z^2$ shows that the existence of one solution implies the existence of a smaller solution ad infinitum. So, there is no such primitive triangle. 
The general case
We require $\lambda(m^2+n^2)$ and $2\lambda m(m+n)$ to be squares and then$$8m(m+n)(m^2+n^2)\text { is a square}.$$
Let $x=\frac{2n}{m}$ then $x^3+2x^2+4x+8$ is a rational square. However
$$y^2=x^3+2x^2+4x+8$$
is an elliptic curve which according to CoCalc has rank $0$ with $(-2,0)$ as the only rational point.
