Suppose that $y$ and $z$ are not co-primes. In that case there must be a prime number p such that $p|y$ and $p|z$. From the first equation we get that $p|x$ and therefore $p|(x+y+z)$ or $p|61$ which is impossible (61 is a prime).
In other words $y$ and $z$ are co-primes. But their product is a perfect square and that is possible only if both numbers are squares. Which means that possible values for $y,z$ are:
$$y,z \in \{1, 4, 9, 16, 25, 36, 49\}$$
...which means that you have just a few pairs to check manualy and most of them can be easily discarded. Also note that at least one of them must be odd (otherwise $x,y,z$ would be all even which is impossible).
For example, 49 can be paired only with 9, 4, 1 (no solution there). 36 can be paired only with 1 and 9 (no solution). 25 can be paired only with 25, 16, 9, 4, 1. We have one solution there ($y=25, z=16, x=20$). Values 16,9,4,1 are just too small (less than 61/3) and should not be checked at all.
So we have exactly two solutions: $y=25, z=16, x=20$ and $y=16, z=25, x=20$