I did find the solution on Internet here.
But I am not able to understand it completely.
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1It doesn't say that $x,y,z$ are integers – Interstigation May 22 '21 at 17:25
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Umm... $x=3$, $y = z = 1$ so $(x + y + z) = 5$. (Alternatively, $x=1$, $y=3$ and $z=1$.)
We can derive this from $x^2 y^5 z = 3^2$ and obvious logic and the prime number theorem (unique prime factorization of integers). Specifically, the right-hand-side has a factorization of $3^2 k$ for some unknown $k \in \mathbb{Z}^+$. Thus by the prime number theorem the left-hand-side must have a factor of the form $3^2$. This can be achieved with $x=3$, or $y=3$ or $z = 9$ (having the smallest candidate contributions to $x+y+z$). Can you continue?
If negative integers are allowed—and they shouldn't be—well then we can make as large a negative number as we want by letting $x = (-9)^{2q+1}$ for $q \to \infty$ and $q \in \mathbb{Z}^+$.
If $\{ x, y, z\} \in \mathbb{R}^+$—and they shouldn't be—then we confront the possibility of $x = z = 1$ and $y^5 = 3^2$ or $y = 3^{2/5} \approx 1.55185$, so $x + y + z = 3.55185$.
David G. Stork
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Negative numbers allowed? Then $x^2 = (-9)^{2 k}$ where $k \to \infty$. – David G. Stork May 22 '21 at 17:15
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1Even if $x,y,z$ are required to be non-negative integers, we can take $x=y=z=0$. – Andreas Blass May 22 '21 at 17:31
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What you (implcitly) use is not normally called the "prime number theorem". Given the OP's difficulty understanding, an answer should be much more explicit with justification. – Bill Dubuque May 22 '21 at 22:26