Clearly $2|n$ $5^m+n^2=3^p\Rightarrow 1+0=(-1)^p\pmod4\Leftrightarrow 2|p$.Let $p=2q$ $5^m=(3^q-n)(3^q+n)$.Let $gcd(3^q-n,3^q+n)=d$.Then $d|2\cdot 3^q\Rightarrow (d,5)=1$ so since $3^q+n>3^q-n$ the only case is $3^q-n=1,3^q+1=5^m\Rightarrow 2\cdot3^q=5^m+1$. If $m\geq 2$ then by Zsigmondy's theorem there exist a prime $p$ such that $p\mid 5^m+1$ and $p\not |5+1=6\Leftrightarrow p\neq 2,3$ contradiction since the only primes divisors of $5^m+1=2\cdot 3^q$ is the $2,3$ So $m=1\Leftrightarrow q=1$ and $n=2$. Therefore the only solution is $\boxed{(m,n,p)=(1,2,2)}$

It's fully similar to Pr0d55's sol, but I was too sad not posting mine

We claim that the only triple of solutions is $(x,y,z)=(2,1,2)$. It is not hard to see that these solutions work, now we shall show that they are the only ones. We start off by considering the given equation $\pmod{4}$. Note that then, \[(-1)^x \equiv 1 + z^2 \pmod{4}\]The left hand side is either $1$ or $-1$ while the right hand side is either $1$ or $2$. Thus, we must have both being $1 \pmod{4}$ with $x$ and $z$ being even. Now that $2\mid x$ we let $x=2r$ for some positive integer $r$. Note that, \begin{align*} 3^x &= 5^y + z^2\\ 3^{2r} &= 5^y + z^2\\ 5^y &= 3^{2r} - z^2\\ 5^y &= (3^r - z)(3^r + z) \end{align*}Note that if $5\mid 3^r -z$ and $5\mid 3^r +z$ we must have $5\mid 2\cdot 3^r$ which is a clear contradiction. Thus, $5$ is not a factor of both these terms. This implies the only possibility is $3^r -z =1$ and $3^r + z = 5^y$. Next, considering the given equation $\pmod{3}$ we have, \[2^y \equiv -z^2\]since the left hand side is 1 or 2 and the right hand side is 0 or 1, we msut have both sides being $1\pmod{3}$ with $y$ being odd. Now, summing the two previous equations we have, \[5^y + 1= 2\cdot 3^r\]By the Exponent Lifting Lemma, \[r = \nu_3(2\cdot 3^r)= \nu_3(5^y+1)= \nu_3(6)+\nu_3(y)=\nu_3(y)+1\]Thus, $\nu_3(y)=r-1$ and $y \ge 3^{r-1}$. But then, \[2\cdot 3^r = 5^y +1 \ge 5^{3^{r-1}}+1 > 3^{3^{r-1}}\]which is clearly false for all $r\ge 2$. Thus, the only possibility is to have $r=1$. This implies $x=2$. Thus, \[5^y + z^2 =9\]which due to size reasons implies $y = 1$. Thus, $z=2$ which implies that our claimed solution is indeed the only one.

Claim: $x \equiv z \equiv 0(\text{mod }2)$ Proof: Consider the equation $\text{modulo } 4$. $$3^x-z^2\equiv 5^y \equiv 1(\text{mod } 4)$$$$ \begin{aligned}2\nmid 3^x &\implies z^2 \equiv 0(\text{mod } 4) \implies z\equiv 0(\text{mod }2)\\ &\implies 3^x \equiv 1(\text{mod } 4) \implies x \equiv 0(\text{mod }2) \end{aligned}$$ Hence, let $x = 2x_1$ and $z = 2z_1$ for some $x_1, z_1 \in \mathbb{Z^{+}}$. Now the original equation simplifies as follows $$3^{2x_1} = 5^y + (2z_1)^2 $$$$(3^{x_1})^2 - (2z_1)^2 = 5^y $$$$ (3^{x_1} - 2z_1)(3^{x_1} + 2z_1) = 5^y$$Hence, $$(3^{x_1} + 2z_1) = 5^m$$$$(3^{x_1} - 2z_1) = 5^n$$for some $m,n \in \mathbb{Z^*}$ such that $y = m+n$ Subtracting the two equations we get $$ 4z_1 = 5^n(5^{m-n} - 1) \implies z_1 = 5^n$$For the sake of contradiction assume $n \geq 1$ which implies that $5 | z$ $$5 | 5^y+z^2 \implies 5 | 3^x$$which is a contradiction. Therefore $n=0 \implies z = 2, y = 1, x = 1 $. Hence the only solution is $(x,y,z) = (2,1,2)$