jmc

Let $x$ be the length of the hypotenuse, and let $y$ be the length of the other leg. Then we have $x^2-y^2=162^2$. Factoring both sides gives $(x+y)(x-y)=(2\times 3^4{)}^2=2^2\times 3^8$. A pair of positive integers $(x,y)$ gives a solution to this equation if and only if $(x+y)$ and $(x-y)$ are factors whose product is $2^2\ast 3^8$. For positive integers $a$ and $b$, the equations $x+y=a$ and $x-y=b$ have positive integer solutions if and only if $a-b$ is an even positive integer. Thus if $ab=2^2\ast 3^8$ and the difference between $a$ and $b$ is even, then we get a valid triangle with $x+y=a$ and $x-y=b$. Since $ab$ is even, at least one of the factors is even, and since their difference is even, the other must be as well. Since $x+y>x-y$ we have $a>b$ i.e. $a>2\times 3^4.$ Since the prime factorization of $a$ must have exactly one $2$, the choices for $a$ that give valid triangles are $2\times 3^5,2\times 3^6,2\times 3^7,2\times 3^8.$ Thus there are $\boxed{4}$ valid triangles.